Polynomials - Class 10 Mathematics - Chapter 2 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Polynomials | NCERT | Mathematics | Class 10
If $\alpha$, $\beta$, and $y$ are the roots of $x^{3} + 8 = 0$, then the equation whose roots are $\alpha^{2}$, $\beta^{2}$, and $y^{2}$ is
(A) $x^{3} - 8 = 0$ (B) $x^{3} - 16 = 0$ (C) $x^{3} + 64 = 0$ (D) $x^{3} - 64 = 0$.
The correct answer is (D) $x^{3} - 64 = 0$.
Given the equation: $$ x^{3} + 8 = 0 $$ where $\alpha$, $\beta$, and $y$ are roots, we want to find the equation with roots squared: $\alpha^{2}$, $\beta^{2}$, and $y^{2}$.
We start by substituting $x = \sqrt{y}$ into the original equation: $$ (\sqrt{y})^3 + 8 = 0 \implies y^{\frac{3}{2}} + 8 = 0 $$
Now, solving for $y^{\frac{3}{2}}$: $$ y^{\frac{3}{2}} = -8 \implies y^3 = (-8)^2 = 64 $$
Therefore, the equation $y^3 - 64 = 0$ or expressed in terms of $x$: $$ x^3 - 64 = 0 $$
This equation has roots $\alpha^2$, $\beta^2$, and $y^2$. Thus, the correct choice is (D) $x^{3} - 64 = 0$.
Which of the following is a binomial?
A) $x^{2} - x + 1$
B) $3x^{2} + 2x^{2} - x^{2}$
C) $\left(x^{2} - 2x\right) x^{-1}$
D) $2x^{2} + x^{2} - 2x + 3 - 2$
Option C $(x^{2} - 2x) x^{-1}$ is the correct choice for a binomial.
A binomial is defined as a polynomial which has exactly 2 terms.
In option A, $x^{2} - x + 1$ consists of 3 terms.
Option B simplifies to $3x^{2} + 2x^{2} - x^{2} = 4x^{2}$, which only has 1 term.
Option C, $(x^{2} - 2x) x^{-1}$, simplifies to $x - 2$, which clearly has 2 terms, satisfying the condition for a binomial.
Option D, $2x^{2} + x^{2} - 2x + 3 - 2$ simplifies to $3x^{2} - 2x + 1$, again containing 3 terms.
Thus, the binomial amongst the options is Option C, $(x^{2} - 2x) x^{-1}$.
The degree of the polynomial $2 - 8x^{0}yz^{3} + xy + 4xyz$ is
A) 1
B) 2
C) 3
D) 4
The correct answer is D) 4.
The polynomial given is:
$$ 2 - 8x^{0}yz^{3} + xy + 4xyz $$
Breaking down the polynomial into its constituent terms:
$2$ has a degree of 0 (since there are no variables).
$-8x^{0}yz^{3}$ simplifies to $-8yz^{3}$ since $x^{0} = 1$. The degree of this term is $1 + 3 = 4$.
$xy$ has the variables x and y both raised to the power of 1. Thus, the degree is $1 + 1 = 2$.
$4xyz$ consists of x, y, and z each raised to the power of 1. Hence, the degree is $1 + 1 + 1 = 3$.
Among the degrees calculated (0, 4, 2, 3), the highest degree is 4.
Thus, the degree of the polynomial is 4.
Zero of the zero degree polynomial is
A. 1
B. 0
C. Any real number
D. Not defined
The correct option is D. Not defined.
The zero of a zero degree polynomial is undefined because any constant non-zero polynomial doesn't have roots where the polynomial equals zero, whereas a zero polynomial (which equals zero for all $x$) doesn't have a specific zero due to having an infinite number of zeros.
The number of cubic polynomials $P(x)$ satisfying $P(1)=2$, $P(2)=4$, $P(3)=6$, $P(4)=8$ is
A. 0
B. 1
C. more than one but finitely many
D. infinitely many
The problem posed requires finding a cubic polynomial $P(x)$ that meets the conditions $P(1)=2$, $P(2)=4$, $P(3)=6$, and $P(4)=8$. A critical observation here is noting the linearity of the conditions given where the outputs change consistently by the same amount as $x$ increments by 1.
Let's begin by observing a simple linear function, $f(x) = 2x$. This function satisfies all the given values: $$ f(1) = 2 \times 1 = 2, \ f(2) = 2 \times 2 = 4, \ f(3) = 2 \times 3 = 6, \ f(4) = 2 \times 4 = 8. $$
Here, $f(x)$ perfectly satisfies all given points. Since $P(x)$ and $f(x)$ intersect at four distinct points namely $x = 1, 2, 3, 4$, and because $f(x)$ is a linear polynomial, there are potentially severe restrictions on what $P(x)$ can be.
The crucial insight here is realizing that if $P(x)$ were a cubic polynomial different from a linear function, it could at most intersect a linear polynomial at three points due to its nature (the Fundamental Theorem of Algebra states a polynomial of degree $n$ has exactly $n$ roots, counting multiplicities). However, since $f(x)$ and $P(x)$ intersect at four points, and $P(x)$ is supposed to be cubic, it would mean $P(x)$ must be identical to $f(x)$, which contradicts the condition that $P(x)$ is cubic.
In conclusion, there is no cubic polynomial that can meet these conditions, and the linear function $f(x) = 2x$ is not cubic but linear. Hence, the number of such cubic polynomials $P(x)$ is: $$ \textbf{0}. $$
The correct option is therefore A. 0.
What must be subtracted from $4x^{3} - 3x + 5$ to get $2x^{2} - 3x^{3} + 5x - 2$?
A) $x^{3} - 2x^{2} - 8x + 7$
B) $7x^{3} + 2x^{2} - 8x + 7$
C) $7x^{3} - 2x^{2} - 8x + 3$
D) $7x^{3} - 2x^{2} - 8x + 7$
To find the polynomial that needs to be subtracted from $4x^{3} - 3x + 5$ to yield $2x^{2} - 3x^{3} + 5x - 2$, we set up the equation: $$ (4x^3 - 3x + 5) - P(x) = 2x^2 - 3x^3 + 5x - 2 $$ where $P(x)$ is the polynomial we need to find.
Rearranging the equation to solve for $P(x)$ gives: $$ P(x) = (4x^3 - 3x + 5) - (2x^2 - 3x^3 + 5x - 2) $$
To find $P(x)$, perform the subtraction operation term by term:
The $x^3$ term: $$ 4x^3 - (-3x^3) = 4x^3 + 3x^3 = 7x^3 $$
The $x^2$ term: $$ 0x^2 - 2x^2 = -2x^2 $$
The $x$ term: $$ -3x - 5x = -8x $$
The constant term: $$ 5 - (-2) = 5 + 2 = 7 $$
Hence, the polynomial $P(x)$ is: $$ 7x^3 - 2x^2 - 8x + 7 $$
Therefore, the correct option is D) $7x^3 - 2x^2 - 8x + 7$.
When $x^{4} + x^{3} - 2x^{2} + x + 1$ is divided by $x - 1$, the remainder is 2 and the quotient is $q(x)$. Find $q(x)$.
A) $x^{3} + 2x^{2} + x + 1$
B) $x^{3} + x^{2} + x$
C) $x^{3} + 2x^{2} + 1$
D) $x^{3} + x^{2} + 1"
The correct answer is C) $$ x^3 + 2x^2 + 1 $$
Given Problem: We are given the polynomial $x^4 + x^3 - 2x^2 + x + 1$. When this polynomial is divided by $x-1$, the remainder is 2 and we need to ascertain the quotient $q(x)$.
Calculation: Using polynomial division, the problem can be set up as: $$ x^4 + x^3 - 2x^2 + x + 1 = (x-1) q(x) + 2 $$ To solve for $q(x)$, you rearrange and simplify: $$ q(x) = \frac{x^4 + x^3 - 2x^2 + x + 1 - 2}{x-1} = \frac{x^4 + x^3 - 2x^2 + x - 1}{x-1} $$ Applying polynomial long division here, you essentially find that the effective terms end up properly dividing to give: $$ x^3 + 2x^2 + 1 $$
Thus, C) $x^3 + 2x^2 + 1$ is the quotient, $q(x)$, when $x^4 + x^3 - 2x^2 + x + 1$ is divided by $x-1$.
The highest index of the variable $x$ occurring in the polynomial $f(x) = a_{0} x^{n} + a_{1} x^{n-1} + a_{2} x^{n-2} + \ldots + a_{n-1} x + a_{n}$ is called _______.
A) Order of the polynomial B) Degree of the polynomial C) Both of these D) None of these
Solution
The correct option is $\mathbf{B}$ Degree of the polynomial
In a polynomial, $$ f(x) = a_0x^n + a_1x^{n-1} + a_2x^{n-2} + \ldots + a_{n-1}x + a_n, $$ the highest power/exponent of the variable $x$ is known as the degree of the polynomial. For the polynomial given in the question, the degree is clearly $n$, which is the power of $x$ in the term $a_0x^n$. Thus, the term describing the highest power of $x$ in a polynomial is the degree of the polynomial.
Degree of a zero polynomial is:
A) 1 B) Undefined C) 0
The correct answer to the question "What is the degree of a zero polynomial?" is:
B) Undefined
The degree of a zero polynomial is categorically undefined. This is because the polynomial $0$, which represents a zero polynomial, does not have a degree that can be concretely determined. Regardless of the power to which the variable is raised (be it $0 \times x^0$, $0 \times x^1$, or $0 \times x^9$), the result always yields the value zero, thus making the determination of its degree impossible.
If $x=0$ and $x=-1$ are the roots of the polynomial $f(x)=2x^{3}-3x^{2}+ax+b$, find the value of $a$ and $b.
Solution:
The polynomial given is: $$ f(x) = 2x^3 - 3x^2 + ax + b $$ Given that the roots of the polynomial include $x = 0$ and $x = -1$. Therefore, $f(0) = 0$ and $f(-1) = 0$ must hold true.
First, evaluating at $x = 0$: $$ f(0) = 2(0)^3 - 3(0)^2 + a \cdot 0 + b = 0 \ 0 - 0 + 0 + b = 0 \ b = 0 $$
With $b = 0$, substitute this value back into the polynomial and evaluate at $x = -1$: $$ f(-1) = 2(-1)^3 - 3(-1)^2 + a \cdot (-1) + 0 = 0 \ 2(-1) - 3 + a \cdot (-1) = 0 \ -2 - 3 - a = 0 \ -a - 5 = 0 \ a = -5 $$
Hence, the coefficients are: $$ a = -5 \quad \text{and} \quad b = 0 $$
The zero of the polynomial $p(x) = 4x + 3$ lies at what value of $x$?
A) 1
B) $\frac{3}{4}$
C) $-\frac{4}{3}$
D) $-\frac{3}{4}$
The correct answer is D) $-\frac{3}{4}$.
To find the zero of the polynomial $p(x) = 4x + 3$, you set the polynomial equal to zero. This zeros represent values of $x$ for which the polynomial equals 0, essentially the points where the graph of the polynomial crosses or touches the $x$-axis.
Thus, solve for $x$ from: $$ 4x + 3 = 0 $$ Subtract 3 from both sides: $$ 4x = -3 $$ Divide by 4: $$ x = -\frac{3}{4} $$
This solution tells us that the zero of the polynomial $p(x) = 4x + 3$ occurs at $x = -\frac{3}{4}$.
13
By actual division, find the quotient and the remainder when the first polynomial $x^{4} + 1$ is divided by the polynomial $(x - 1)$.
To solve the problem of dividing the polynomial ( x^4 + 1 ) by ( (x-1) ), we use the long division method in polynomials. The steps in long division help us iteratively reduce the problem until we derive both the quotient and the remainder.
Here's a breakdown after performing the division step-by-step:
- First, we determine how many times ( (x-1) ) fits into the leading term of ( x^4 + 1 ), which is ( x^4 ). We see that ( x^3 ) (since ( x^3 \cdot (x-1) = x^4 - x^3 )) will be the first term of our quotient.
- We subtract ( x^4 - x^3 ) from ( x^4 + 1 ), which eliminates the ( x^4 ) term and leaves us with ( x^3 + 1 ).
- We repeat this process for each power of ( x ) until reaching the constant term, iterating through ( x^2 ), ( x ), and 1 as the subsequent terms in the quotient.
- This sequence gives us a quotient of ( x^3 + x^2 + x + 1 ) and finally a remainder of 2 (as there is no further ( x ) term to go into from 2).
Therefore, after performing the actual polynomial division of ( x^4 + 1 ) by ( x-1 ), the quotient is ( \mathbf{x^3 + x^2 + x + 1} ) and the remainder is ( \mathbf{2} ).
Total number of polynomials of the form $x^3 + ax^2 + bx + c$, that are divisible by $x^2 + 1$, where $a, b, c \in {1, 2, 3, \ldots, 9, 10}$ is
A) 8
B) 10
C) $\mathbf{15}$
D) 30
The correct answer is Option B: 10.
For the polynomial $$p(x) = x^3 + ax^2 + bx + c,$$ to be divisible by the polynomial $$x^2 + 1 = (x-i)(x+i),$$ where $i$ is the imaginary unit, we require that both $p(i) = 0$ and $p(-i) = 0$.
Calculating $p(i)$, we get: $$ i^3 + ai^2 + bi + c = -i + a(-1) + bi + c = 0. $$ This simplifies to: $$ (c - a) + (b - 1)i = 0. $$
Similarly, for $p(-i)$: $$ (-i)^3 + a(-i)^2 + b(-i) + c = i + a(-1) - bi + c = 0. $$ This simplifies to: $$ (c - a) + (1 - b)i = 0. $$
Both of these equations must hold true. We get:
- From the real parts: $c - a = 0 \implies c = a$,
- From the imaginary parts: $b - 1 = 0 \implies b = 1$.
Thus, $a=c$ and $b=1$.
Given that $a, c \in {1, 2, 3, ..., 10}$, $a$ and $c$ can be any of the 10 given integers, as long as they are equal. With $b$ fixed as 1, each combination of $a = c$ yields a valid polynomial.
Therefore, the total number of such polynomials is $$\binom{10}{1} = 10.$$
When $p(x)$ is divided by $x - ab$, the remainder is
A) $p(a+b)$
B) $p(a)$
C) $p(ab)$
D) $p(b)$
The remainder when a polynomial $p(x)$ is divided by $x - r$ is given by $p(r)$. Here, we are dividing by $x - ab$. Using this rule, the remainder is $p(ab)$.
Correct answer: C) $p(ab)$
What remains when $7x - 3x^{2}$ is taken away from $4x + 8x^{2}$?
A) $-3x$
B) $11x^{2} - 5x$
C) $11x^{2}$
D) $11x^{2} - 3x$
To solve the problem, we need to subtract the expression $7x - 3x^2$ from $4x + 8x^2$. Let's go through the steps:
-
Write down the expressions: $$ (4x + 8x^2) - (7x - 3x^2) $$
-
Remove the parentheses. Remember to change the signs within the parentheses that is being subtracted: $$ 4x + 8x^2 - 7x + 3x^2 $$
-
Combine like terms. Gather all $x^2$ terms and all $x$ terms: $$ (8x^2 + 3x^2) + (4x - 7x) = 11x^2 - 3x $$
Therefore, after subtracting $7x - 3x^2$ from $4x + 8x^2$, the result is $11x^2 - 3x$.
The correct option is D.
If $f(x)f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \forall x \in \mathbb{R}-{0}$ where $f(x)$ is a polynomial function and $f(5) = 126$, then $f(3) =$
A) 28
B) 26
C) 27
D) 25
Solution:
Given the functional equation: $$ f(x)f\left(\frac{1}{x}\right) = f(x) + f\left(\frac{1}{x}\right) \quad \text{for all } x \in \mathbb{R} \setminus {0} $$ where $ f(x) $ is a polynomial function. Rewriting the equation, we can define it as: $$ f(x)f\left(\frac{1}{x}\right) - f(x) - f\left(\frac{1}{x}\right) = 0 $$ This implies that $f(x)$ and $f\left(\frac{1}{x}\right)$ must have specific factors that allow this equation to be true for all $x \neq 0$. A simple candidate that fits into both polynomial and functional properties is: $$ f(x) = 1 \pm x^n $$ Given $f(5) = 126$, substituting in this form, we have: $$ 1 \pm 5^n = 126 $$ Solving for $5^n$, we get: $$ \pm 5^n = 125 \quad \text{or} \quad 5^n = 125 $$ Since $125 = 5^3$, this tells us that $n = 3$ and the correct form of $f(x)$ should ensure this equality works: $$ f(x) = 1 + x^3 $$ To find $f(3)$: $$ f(3) = 1 + 3^3 = 1 + 27 = 28 $$ Therefore, the value of $f(3)$ is 28. Thus, the correct option is:
A) 28
Let $a$ be the remainder when $p(x)=4x+8$ is divided by $(x-2)$. What is the value of $(a+2)$?
A) 16
B) 18
C) 20
D) 22
Using the remainder theorem, which states that the remainder of the polynomial $p(x)$ divided by $(x-a)$ is $p(a)$, we can find the remainder when $p(x) = 4x + 8$ is divided by $(x-2)$.
Set: $$ p(x) = 4x + 8 $$ Divisor: $$ x - 2 = 0 \Rightarrow x = 2 $$ By substituting $x = 2$ into $p(x)$, we get the remainder (denoted as $a$): $$ p(2) = 4(2) + 8 = 8 + 8 = 16 $$ Thus, $a = 16$, and we need to find the value of $a + 2$: $$ a + 2 = 16 + 2 = 18 $$ Therefore, the value of $(a + 2)$ is 18. The correct option is B) 18.
Let $f(x)$ be a polynomial of degree 5 such that $x = \pm 1$ are its critical points. If $\lim_{x \to 0} \left(2 + \frac{f(x)}{x^{3}}\right) = 4$, then which of the following is not true?
A $f(1) - 4 f(-1) = 4$.
B $x = 1$ is a point of maxima and $x = -1$ is a point of minimum of $f$.
C $f$ is an odd function.
D $x = 1$ is a point of minima and $x = -1$ is a point of maxima of $f$.
The correct option is D: $x = 1$ is a point of minima and $x = -1$ is a point of maxima of $f$.
Given the condition: $$ \lim_{x \to 0} \left(2 + \frac{f(x)}{x^3}\right) = 4 $$ this implies: $$ \lim_{x \to 0} \frac{f(x)}{x^3} = 2 $$ Since the limit is finite, let $f(x)$ be of the form: $$ f(x) = a x^5 + b x^4 + c x^3 $$ Considering the limit: $$ \lim_{x \to 0} (a x^2 + b x + c) = 2 $$ we get: $$ c = 2 $$
Furthermore, the derivative $f'(x) = 5ax^4 + 4bx^3 + 3cx^2$ at critical points $x = \pm 1$ gives: $$ f'(1) = 5a + 4b + 3 \cdot 2 = 0 \ f'(-1) = 5a - 4b + 3 \cdot 2 = 0 $$ By solving these equations, we find: $$ b = 0, \quad a = -\frac{6}{5} $$ Thus: $$ f(x) = -\frac{6}{5} x^5 + 2x^3 $$ From here, we know $f(x)$ is an odd function.
We now check the second derivative to test local maxima or minima conditions: $$ f''(x) = -24x^3 + 12x $$ Evaluating at $x = \pm 1$, we find: $$ f''(1) = -12 < 0 \text{ (indicates local maxima)} \ f''(-1) = 12 > 0 \text{ (indicates local minima)} $$
Finally, using the formula for $f(x)$, checking the value: $$ f(1) - 4f(-1) = 4 - (-24) = 4 $$ Confirming Option A is true.
Since we have proven $x = 1$ is a maxima and $x = -1$ is a minima, Option D is not true, confirming it as the correct choice.
The degree of the polynomial $3z^{2} + z + 1$ is 2.
A) True
B) False
The solution provided is incorrect in determining the truth value and here is the correct evaluation:
Option: A
True
The polynomial provided is: $$ 3z^2 + z + 1 $$
To find the degree of a polynomial, we must identify the highest exponent of the variable in the polynomial. Here, $z$ is the variable with terms:
- $3z^2$ (where the exponent of $z$ is 2),
- $z$ (where the exponent of $z$ is 1),
- $1$ (a constant term with an implicit exponent of 0 for $z$).
The highest exponent among the terms is 2 (from the term $3z^2$). Therefore, the degree of this polynomial is indeed 2.
As the statement asserts the degree of the polynomial $3z^2 + z + 1$ is 2, and we have confirmed this to be true, the correct answer should be True. The provided solution incorrectly judged the statement to be false.
3 Verify whether the following are zeroes of the polynomial, indicated against them. (i) $p(x) = 3x + 1$, $x = -\frac{1}{3}$ (ii) $p(x) = 5x - \pi$, $x = \frac{4}{5}$ (iii) $p(x) = x^2 - 1$, $x = 1, -1$ (iv) $p(x) = (x + 1)(x - 2)$, $x = -1, 2$ (v) $p(x) = x^2$, $x = 0$ (vi) $p(x) = ix + m$, $x = -\frac{m}{i}$ (vii) $p(x) = 3x^2 - 1$, $x = -\frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}}$ (viii) $p(x) = 2x + 1$, $x = \frac{1}{2}`
Solution
(i) Yes, $-\frac{1}{3}$ is the zero of the polynomial. We calculate: $$ p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0 $$ Thus, $x = -\frac{1}{3}$ is indeed a zero of $p(x) = 3x + 1$.
(ii) No, $x = \frac{4}{5}$ is not the zero of the polynomial. We find: $$ p(\frac{4}{5}) = 5(\frac{4}{5}) - \pi = 4 - \pi \neq 0 $$ Therefore, $x = \frac{4}{5}$ does not satisfy the polynomial $p(x) = 5x - \pi$.
(iii) Yes, $x = 1$ and $x = -1$ are zeros of the polynomial. We compute: $$ p(1) = 1^2 - 1 = 0 \quad \text{and} \quad p(-1) = (-1)^2 - 1 = 0 $$ Both $x = 1$ and $x = -1$ are zeros of $p(x) = x^2 - 1$.
(iv) Yes, $x = -1$ and $x = 2$ are zeros of the polynomial since: $$ p(-1) = (-1+1)(-1-2) = 0 \quad \text{and} \quad p(2) = (2+1)(2-2) = 0 $$ Therefore, both $x = -1$ and $x = 2$ satisfy $p(x) = (x + 1)(x - 2)$.
(v) Yes, $x = 0$ is the zero of the polynomial. Calculating: $$ p(0) = 0^2 = 0 $$ Hence, $x = 0$ is a zero of $p(x) = x^2$.
(vi) Yes, $x = -\frac{m}{i}$ is the zero of the polynomial. Assume $x + m = 0$, then: $$ i(-\frac{m}{i}) + m = 0 \quad \text{(simplified as $-m + m = 0$)} $$
(vii) Yes, $x = -\frac{1}{\sqrt{3}}$ is a zero, but $x = \frac{2}{\sqrt{3}}$ is not. Verifying: $$ p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 - 1 = 1 - 1 = 0 \quad \text{and} \ p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 - 1 = 4 - 1 = 3 \neq 0 $$ Only $x = -\frac{1}{\sqrt{3}}$ is a zero of $p(x) = 3x^2 - 1$.
(viii) No, $x = \frac{1}{2}$ is not the zero of the polynomial. We confirm: $$ p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2 \neq 0 $$ Thus, $x = \frac{1}{2}$ does not satisfy $p(x) = 2x + 1$.
Divide $\left(2x^{3} - 3x^{2} - 10x + 5\right)$ by $(2x - 3)$ and write the quotient and the remainder.
To divide the polynomial $ 2x^{3} - 3x^{2} - 10x + 5 $$ by $$ 2x - 3 $ and find the quotient and remainder, we perform polynomial long division.
Determining the first term of the quotient:Divide the leading term of the numerator, $2x^3$, by the leading term of the denominator, $2x$. This gives: $$ \frac{2x^3}{2x} = x^2 $$ So, the first term of our quotient is $x^2$.
Multiply and subtract:Multiply $x^2$ by the whole divisor $(2x - 3)$: $$ x^2 \cdot (2x - 3) = 2x^3 - 3x^2 $$ Subtract this from the original polynomial: $$ (2x^3 - 3x^2 - 10x + 5) - (2x^3 - 3x^2) = -10x + 5 $$
Determining the next term of the quotient:Again, divide the leading term of the new result, $-10x$, by the leading term of the divisor, $2x$: $$ \frac{-10x}{2x} = -5 $$ This is the next term of the quotient.
Multiply and subtract again:Multiply $-5$ by the divisor $(2x - 3)$: $$ -5 \cdot (2x - 3) = -10x + 15 $$ Subtract this from the previous result: $$ (-10x + 5) - (-10x + 15) = 5 - 15 = -10 $$ This is now our remainder, as there are no more terms with $x$ in the divisor to continue division.
Stating the results:The quotient from the division is $$ x^2 - 5 $$ and the remainder is $$ -10 $$.
Thus, when dividing $\left(2x^{3} - 3x^{2} - 10x + 5\right)$ by $(2x - 3)$, the quotient is $x^2 - 5$ and the remainder is $-10$.
Form a cubic polynomial in $x$ with the sum, sum of the products of its zeros taken two at a time, and product of its zeros are 8, 0, and 9 respectively.
To form a cubic polynomial in $x$ with the given properties of its zeros (roots), let's denote the zeros of the polynomial as $\alpha$, $\beta$, and $\gamma$. According to Vieta's formulas for roots of polynomials, the following relationships hold for a cubic polynomial $ax^3 + bx^2 + cx + d$:
The sum of the zeros ($\alpha + \beta + \gamma$) is equal to $-b/a$.
The sum of the products of the zeros taken two at a time ($\alpha \beta + \beta \gamma + \gamma \alpha$) is equal to $c/a$.
The product of the zeros ($\alpha \beta \gamma$) is equal to $-d/a$.
For our specific case:
Given $\alpha + \beta + \gamma = 8$,
$\alpha \beta + \beta \gamma + \gamma \alpha = 0$,
$\alpha \beta \gamma = 9$.
We need to form a polynomial where $a = 1$ (generally for simplicity and convention in forming such polynomials). Plugging into the formulas:
The sum of the zeros (coefficient of $x^2$) should be $-(-8) = 8$.
The sum of the products of the zeros taken two at a time (coefficient of $x$) should be $0$.
The product of the zeros (constant term) should be $-9$.
Thus, the polynomial is formed by integrating these values: $$ x^3 - 8x^2 + 0x - 9 = 0 $$
Simplifying further, we yield the final polynomial: $$ x^3 - 8x^2 - 9 = 0 $$
And there you have it, a cubic polynomial $x^3 - 8x^2 - 9$ which meets the stated requirements.
Examine if 1 and 2 are zeros of the polynomial $p(x) = x^{3} - 3x^{2} - x + 3$.
To verify if 1 and 2 are zeros of the polynomial $ p(x) = x^3 - 3x^2 - x + 3 $, we’ll evaluate $ p(x) $ at these values and check if each result equals zero.
Testing for $ x = 1 $:
Plug 1 into the polynomial:
$$
p(1) = 1^3 - 3 \times 1^2 - 1 + 3
$$Simplify: $$ p(1) = 1 - 3 - 1 + 3 = 0 $$
Since $ p(1) = 0 $, we confirm that $ x = 1 $ is a zero of the polynomial.
Testing for $x = 2 $:
Plug 2 into the polynomial: $$ p(2) = 2^3 - 3 \times 2^2 - 2 + 3 $$
Simplify: $$ p(2) = 8 - 3 \times 4 - 2 + 3 = 8 - 12 - 2 + 3 = -3 $$
Since $p(2) \neq 0 $ (specifically, $ p(2) = -3 $), $x = 2 $ is not a zero of the polynomial.
Conclusion: $ x = 1 $ is a zero of the polynomial $ p(x) = x^3 - 3x^2 - x + 3 $, but $ x = 2$ is not a zero of this polynomial.
Find the degree of the polynomial: $(x+1)(x^{2}-x+x^{4}-1)$.
To find the degree of the given polynomial $(x+1)(x^2-x+x^4-1)$, we need to multiply and simplify the expression to identify the term with the highest power of $x$. Let's work this through step-by-step.
Expression:
$$ f(x) = (x+1)(x^2 - x + x^4 - 1) $$
Steps for Multiplication:
Multiply $x$ by all terms inside the second bracket: $$ x \cdot x^2 = x^3, \quad x \cdot (-x) = -x^2, \quad x \cdot x^4 = x^5, \quad x \cdot (-1) = -x $$ This simplifies to: $$ x^5 - x^2 + x^3 - x $$
Multiply $1$ by all terms inside the second bracket: $$ 1 \cdot x^2 = x^2, \quad 1 \cdot (-x) = -x, \quad 1 \cdot x^4 = x^4, \quad 1 \cdot (-1) = -1 $$ This gives: $$ x^4 + x^2 - x - 1 $$
Adding the Results from Both Multiplications:
Combine like terms from the two products: $$ x^5 - x^2 + x^3 - x + x^4 + x^2 - x - 1 $$
Simplify by combining like terms: $$ x^5 + x^4 + x^3 - x^2 - x^2 - x - x - 1 $$
Further simplification results in: $$ x^5 + x^4 + x^3 - 2x - 1 $$
Determination of the Degree:
The degree of a polynomial is the highest power of $x$ present in the expression once it's been fully simplified. From our expression $x^5 + x^4 + x^3 - 2x - 1$, the highest power of $x$ is $5$.
Conclusion:
The degree of the polynomial $(x+1)(x^2-x+x^4-1)$ is 5.
What should be added to the polynomial $x^{2}-5x+4$, so that 3 is a zero of the resulting polynomial?
(a) 1
(b) 2
(c) 4
(d) 5
To find out what value should be added to the polynomial ( x^2 - 5x + 4 ) such that 3 is a zero of the resulting polynomial, we need to find some value $a$, so when added, 3 satisfies the equation:
$$ (x^2 - 5x + 4) + a = 0. $$ Substituting $x = 3$ into the equation gives: $$ (3^2 - 5 \times 3 + 4) + a = 0 $$ Calculate inside the parentheses: $$ (9 - 15 + 4) + a = 0. $$ This simplifies to: $$ (-2) + a = 0. $$ To solve for $a$, isolate it on one side: $$ a = 2. $$ Thus, adding 2 to the polynomial ( x^2 - 5x + 4 ) will make 3 a zero of the resulting polynomial. The correct answer is (b) 2.
Check whether $g(x)$ is a factor of $p(x)$ by dividing polynomial $p(x)$ by polynomial $g(x)$, where
$$ p(x) = x^5 - 4x^3 + x^2 + 3x + 1, \quad g(x) = x^3 - 3x + 1 $$
To determine whether the polynomial $g(x) = x^3 - 3x + 1$ is a factor of the polynomial $p(x) = x^5 - 4x^3 + x^2 + 3x + 1$, we will perform polynomial division of $p(x)$ by $g(x)$. If the remainder of this division is zero, then $g(x)$ is a factor of $p(x)$.
Steps for Polynomial Division
Write the Dividend and the Divisor:
Dividend: $p(x) = x^5 - 4x^3 + x^2 + 3x + 1$
Divisor: $g(x) = x^3 - 3x + 1$
Divide the Leading Terms:
Divide the leading term of $p(x)$ by the leading term of $g(x)$: $$ \frac{x^5}{x^3} = x^2 $$
Multiply $x^2$ by $g(x)$ and subtract from $p(x)$:
$$ x^2 \cdot g(x) = x^2(x^3 - 3x + 1) = x^5 - 3x^3 + x^2 $$ $$ (x^5 - 4x^3 + x^2 + 3x + 1) - (x^5 - 3x^3 + x^2) = -x^3 + 3x + 1 $$
Repeat the Process for the New Dividend ($-x^3 + 3x + 1$):
Divide $-x^3$ by $x^3$: $$ \frac{-x^3}{x^3} = -1 $$
Multiply $-1$ by $g(x)$ and subtract:
$$ -1 \cdot g(x) = -1(x^3 - 3x + 1) = -x^3 + 3x - 1 $$ $$ (-x^3 + 3x + 1) - (-x^3 + 3x - 1) = 2 $$
Check the Remainder:
The final remainder is $2$, which is not zero.
Conclusion
Since the remainder from dividing $p(x)$ by $g(x)$ is $2$ and not zero, $g(x)$ is not a factor of $p(x)$. Thus, $g(x)$ fully does not divide $p(x)$ and hence, it cannot be considered a factor of $p(x)$.
Find the remainder when $x^4 + 12x^3 - 5x^2 + 7x - 200$ is divided by $x - 3$.
To find the remainder when $ x^4 + 12x^3 - 5x^2 + 7x - 200 $ is divided by $ x - 3 $, let's use synthetic division.
Let's consider ( p(x) = x^4 + 12x^3 - 5x^2 + 7x - 200 ). The divisor $x - 3$ gives us the value $ x = 3 $.
Using synthetic division method, we follow these steps:
$$ \begin{array}{r|rrrrr} 3 & 1 & 12 & -5 & 7 & -200 \ \hline & & 3 & 45 & 120 & 381 \ \hline & 1 & 15 & 40 & 127 & 181 \ \end{array} $$
Here’s a step-by-step breakdown:
Write down the coefficients of $ p(x) $: $ 1, 12, -5, 7, -200 $.
Bring down the leading coefficient ($ 1 $) to the bottom row.
Multiply this number by $ 3 $ (the root from $ x-3 $) and write the result below the next coefficient.
Add this result to the next coefficient.
Continue this process for all coefficients.
Following this process:
Bring down $1$.
$ 1 \times 3 = 3 $. Add it to $ 12 $ to get $ 15 $.
$ 15 \times 3 = 45 $. Add it to $ -5 $ to get $ 40 $.
$ 40 \times 3 = 120 $. Add it to $ 7 $ to get $ 127 $.
$ 127 \times 3 = 381 $. Add it to $ -200 $ to get $ 181 $.
The remainder, which is the last value in the bottom row, is 181.
Thus, the remainder when $ x^4 + 12x^3 - 5x^2 + 7x - 200 $ is divided by $ x - 3 $ is 181.
The degree of the polynomial obtained when $8-6x+x^2-7x^3+x^5$ is subtracted from $x^4-6x^3+x^2-3x+1$ is:
A) 5
B) 4
C) 3
D) 1
The correct option is A, 5.
We need to find the degree of the polynomial resulting from the subtraction of $8-6x+x^2-7x^3+x^5$ from $x^4-6x^3+x^2-3x+1$.
Let's perform the subtraction: $$ \begin{array}{r} \left(x^{4} - 6x^{3} + x^{2} - 3x + 1\right) \\ \left(x^{5} - 7x^{3} + x^{2} - 6x + 8\right) \end{array} $$
Expanding and simplifying the expression, we get:
$$ \begin{align*} x^4 & - 6x^3 + x^2 - 3x + 1 \ & - x^5 + 7x^3 - x^2 + 6x - 8 \ = & -x^5 + x^4 + x^3 + 3x - 7 \end{align*} $$
The degree of a polynomial is defined as the highest exponent of its terms. The terms of the resulting polynomial are $-x^5$, $x^4$, $x^3$, $3x$, and $-7$.
Among these, the term $-x^5$ has the highest exponent, which is 5. Hence, the degree of the polynomial is 5.
What must be subtracted from $(x^{3}-3x^{2}+5x-1)$ to get $(2x^{3}+x^{2}-4x+2)$?
A $-x^{3}+4x^{2}+9x-3$ B $-x^{3}-4x^{2}+6x-3$ C $-x^{3}-4x^{2}+9x-9$ D $-x^{3}-4x^{2}+9x-3$
The correct option is $\mathbf{D}$: $$ -x^{3} - 4x^{2} + 9x - 3 $$
Let the unknown value to be subtracted be represented as $ y $.
To find $ y $, we start with the equation: $$ \left( x^{3} - 3x^{2} + 5x - 1 \right) - y = 2x^{3} + x^{2} - 4x + 2 $$
Solving for $ y $, we get: $$ y = \left( x^{3} - 3x^{2} + 5x - 1 \right) - \left( 2x^{3} + x^{2} - 4x + 2 \right) $$
This simplifies to: $$ \begin{array}{rl} y = & x^{3} - 3x^{2} + 5x - 1 \ & - (2x^{3} + x^{2} - 4x + 2) \ y = & x^{3} - 3x^{2} + 5x - 1 \ & - 2x^{3} - x^{2} + 4x - 2 \ y = & - x^{3} - 4x^{2} + 9x - 3 \end{array} $$
Thus, the expression that must be subtracted is: $$ \mathbf{-x^{3} - 4x^{2} + 9x - 3} $$
Which of the following algebraic expressions is a trinomial?
Option 1) 2x + 6 + 3x
Option 2) 5y + 15xy + 6x
Option 3) 9x + 2y - 6xy - 9x
Option 4) 7y + 21z - 9x + 6xy
The correct option is B) 5y + 15xy + 6x
Option 1:
$$ 2x + 6 + 3x = 5x + 6 $$
This expression contains two terms, making it a binomial.Option 2:
$$ 5y + 15xy + 6x $$
This expression contains three terms, making it a trinomial.Option 3:
Simplifying $$ 9x + 2y - 6xy - 9x \rightarrow 2y - 6xy $$
This expression contains two terms, making it a binomial.Option 4:
$$ 7y + 21z - 9x + 6xy $$
This expression contains four terms, making it a 4-term polynomial.
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