Zeroes of a Polynomial Questions in English

(A) Let the polynomial be $p(x) = x + 2$.
To check if $x = a$ is a zero of the polynomial,we evaluate $p(a)$. If $p(a) = 0$,then $a$ is a zero.
For $x = 2$:
$p(2) = 2 + 2 = 4$.
Since $p(2) \neq 0$,$2$ is not a zero of the polynomial.
For $x = -2$:
$p(-2) = -2 + 2 = 0$.
Since $p(-2) = 0$,$-2$ is a zero of the polynomial.
Therefore,$-2$ is a zero of the polynomial $x + 2$,but $2$ is not.

(B) To find a zero of the polynomial $p(x)$,we set the polynomial equal to zero:
$p(x) = 0$
$2x + 1 = 0$
Subtract $1$ from both sides:
$2x = -1$
Divide both sides by $2$:
$x = -\frac{1}{2}$
Therefore,$-\frac{1}{2}$ is the zero of the polynomial $p(x) = 2x + 1$.

(A) Let $p(x) = x^{2} - 2x$.
To verify if $2$ is a zero,we calculate $p(2)$:
$p(2) = (2)^{2} - 2(2) = 4 - 4 = 0$.
Since $p(2) = 0$,$2$ is a zero of the polynomial.
To verify if $0$ is a zero,we calculate $p(0)$:
$p(0) = (0)^{2} - 2(0) = 0 - 0 = 0$.
Since $p(0) = 0$,$0$ is a zero of the polynomial.
Hence,both $2$ and $0$ are zeroes of the polynomial $x^{2} - 2x$.

(N/A) Given the polynomial $p(x) = (x - 1)(x + 1)$.
To find $p(0)$,substitute $x = 0$ into the polynomial:
$p(0) = (0 - 1)(0 + 1) = (-1)(1) = -1$.
To find $p(1)$,substitute $x = 1$ into the polynomial:
$p(1) = (1 - 1)(1 + 1) = (0)(2) = 0$.
To find $p(2)$,substitute $x = 2$ into the polynomial:
$p(2) = (2 - 1)(2 + 1) = (1)(3) = 3$.

(A) To verify if $x = -\frac{1}{3}$ is a zero of the polynomial $p(x) = 3x + 1$,we substitute $x = -\frac{1}{3}$ into the polynomial.
$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1$
$p\left(-\frac{1}{3}\right) = -1 + 1$
$p\left(-\frac{1}{3}\right) = 0$
Since the value of the polynomial at $x = -\frac{1}{3}$ is $0$,it is confirmed that $x = -\frac{1}{3}$ is a zero of the given polynomial.

(B) To verify if $x = \frac{4}{5}$ is a zero of the polynomial $p(x) = 5x - \pi$,we substitute $x = \frac{4}{5}$ into the polynomial.
$p\left(\frac{4}{5}\right) = 5\left(\frac{4}{5}\right) - \pi$
$p\left(\frac{4}{5}\right) = 4 - \pi$
Since $4 - \pi \neq 0$,the value of the polynomial at $x = \frac{4}{5}$ is not zero.
Therefore,$x = \frac{4}{5}$ is not a zero of the given polynomial $p(x) = 5x - \pi$.

(A) To verify if $x = 1$ and $x = -1$ are zeroes of the polynomial $p(x) = x^{2} - 1$,we evaluate the polynomial at these values.
For $x = 1$:
$p(1) = (1)^{2} - 1 = 1 - 1 = 0$
For $x = -1$:
$p(-1) = (-1)^{2} - 1 = 1 - 1 = 0$
Since $p(1) = 0$ and $p(-1) = 0$,both $x = 1$ and $x = -1$ are zeroes of the given polynomial $p(x) = x^{2} - 1$.

(N/A) If $x = -1$ and $x = 2$ are zeroes of the polynomial $p(x) = (x + 1)(x - 2)$,then $p(-1)$ and $p(2)$ must be equal to $0$.
First,substitute $x = -1$ into the polynomial:
$p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0$.
Next,substitute $x = 2$ into the polynomial:
$p(2) = (2 + 1)(2 - 2) = (3)(0) = 0$.
Since both $p(-1) = 0$ and $p(2) = 0$,it is verified that $x = -1$ and $x = 2$ are indeed the zeroes of the given polynomial.

(A) To verify if $x = 0$ is a zero of the polynomial $p(x) = x^{2}$,we need to calculate the value of $p(0)$.
Substitute $x = 0$ into the polynomial:
$p(0) = (0)^{2} = 0$.
Since $p(0) = 0$,the value $x = 0$ is indeed a zero of the polynomial $p(x) = x^{2}$.

(A) To verify if $x = -\frac{m}{l}$ is a zero of the polynomial $p(x) = lx + m$,we substitute the value of $x$ into the polynomial.
$p\left(-\frac{m}{l}\right) = l\left(-\frac{m}{l}\right) + m$
$p\left(-\frac{m}{l}\right) = -m + m$
$p\left(-\frac{m}{l}\right) = 0$
Since the value of the polynomial at $x = -\frac{m}{l}$ is $0$,it is confirmed that $x = -\frac{m}{l}$ is a zero of the given polynomial.

(N/A) To verify if the given values are zeroes of the polynomial $p(x) = 3x^2 - 1$,we substitute the values of $x$ into the polynomial. If the result is $0$,then the value is a zero of the polynomial.
Step $1$: Check for $x = -\frac{1}{\sqrt{3}}$:
$p\left(-\frac{1}{\sqrt{3}}\right) = 3\left(-\frac{1}{\sqrt{3}}\right)^2 - 1$
$= 3\left(\frac{1}{3}\right) - 1$
$= 1 - 1 = 0$
Since $p\left(-\frac{1}{\sqrt{3}}\right) = 0$,$x = -\frac{1}{\sqrt{3}}$ is a zero of the polynomial.
Step $2$: Check for $x = \frac{2}{\sqrt{3}}$:
$p\left(\frac{2}{\sqrt{3}}\right) = 3\left(\frac{2}{\sqrt{3}}\right)^2 - 1$
$= 3\left(\frac{4}{3}\right) - 1$
$= 4 - 1 = 3$
Since $p\left(\frac{2}{\sqrt{3}}\right) \neq 0$,$x = \frac{2}{\sqrt{3}}$ is not a zero of the polynomial.

(B) If $x = \frac{1}{2}$ is a zero of the polynomial $p(x) = 2x + 1$,then $p(\frac{1}{2})$ must be equal to $0$.
Substituting $x = \frac{1}{2}$ into the polynomial:
$p(\frac{1}{2}) = 2(\frac{1}{2}) + 1$
$p(\frac{1}{2}) = 1 + 1 = 2$
Since $p(\frac{1}{2}) = 2 \neq 0$,the value $x = \frac{1}{2}$ is not a zero of the given polynomial.

(B) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = x + 5$.
Setting $p(x) = 0$,we get $x + 5 = 0$.
Solving for $x$,we subtract $5$ from both sides: $x = -5$.
Therefore,the zero of the polynomial $p(x) = x + 5$ is $-5$.

(C) To find the zero of the polynomial $p(x) = x - 5$,we set the polynomial equal to zero.
$p(x) = 0$
$x - 5 = 0$
Adding $5$ to both sides,we get:
$x = 5$
Thus,the zero of the polynomial $p(x) = x - 5$ is $5$.

(D) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = 2x + 5$.
Setting the polynomial equal to zero:
$2x + 5 = 0$
Subtract $5$ from both sides:
$2x = -5$
Divide both sides by $2$:
$x = \frac{-5}{2}$
Thus,the zero of the polynomial $p(x) = 2x + 5$ is $\frac{-5}{2}$.

(A) To find the zero of the polynomial $p(x) = 3x - 2$,we set $p(x) = 0$.
$3x - 2 = 0$
Adding $2$ to both sides,we get $3x = 2$.
Dividing both sides by $3$,we get $x = \frac{2}{3}$.
Thus,the zero of the polynomial $p(x) = 3x - 2$ is $\frac{2}{3}$.

(B) To find the zero of the polynomial $p(x) = 3x$,we set $p(x) = 0$.
$3x = 0$
Dividing both sides by $3$,we get:
$x = \frac{0}{3}$
$x = 0$
Thus,the zero of the polynomial $p(x) = 3x$ is $0$.

(C) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = ax$ and $a \neq 0$.
Setting the polynomial equal to zero: $ax = 0$.
Since $a \neq 0$,we can divide both sides by $a$: $x = 0/a$.
Therefore,$x = 0$.
Thus,the zero of the polynomial $p(x) = ax$ is $0$.

(A) To find the zero of the polynomial $p(x) = cx + d$,we set $p(x) = 0$.
$cx + d = 0$
Subtracting $d$ from both sides,we get:
$cx = -d$
Dividing both sides by $c$ (since $c \neq 0$):
$x = -\frac{d}{c}$
Thus,the zero of the polynomial $p(x) = cx + d$ is $-\frac{d}{c}$.

(B) zero polynomial is a polynomial of the form $p(x) = 0$.
By definition,a zero of a polynomial $p(x)$ is a value $c$ such that $p(c) = 0$.
Since $p(x) = 0$ for all real values of $x$,every real number is a zero of the zero polynomial.
However,the degree of the zero polynomial is not defined.
In the context of standard mathematical definitions,the zeros of the zero polynomial are all real numbers.
Therefore,option $(B)$ is the correct answer.

(B) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = 2x + 5$,we set $2x + 5 = 0$.
Subtracting $5$ from both sides,we get $2x = -5$.
Dividing both sides by $2$,we get $x = -\frac{5}{2}$.
Therefore,$-\frac{5}{2}$ is the zero of the polynomial.
Hence,$(b)$ is the correct answer.

(C) Let the polynomial be $p(x) = 2x^2 + 7x - 4$.
To find the zeroes,we set $p(x) = 0$:
$2x^2 + 7x - 4 = 0$
$2x^2 + 8x - x - 4 = 0$
$2x(x + 4) - 1(x + 4) = 0$
$(2x - 1)(x + 4) = 0$
This gives $2x - 1 = 0$ or $x + 4 = 0$.
So,$x = \frac{1}{2}$ or $x = -4$.
Comparing these with the given options,$\frac{1}{2}$ is one of the zeroes.
Therefore,option $(c)$ is correct.

(FALSE) The given statement is False.
Justification: $A$ zero of a polynomial $p(x)$ is a value $k$ such that $p(k) = 0$. For example,consider the polynomial $p(x) = x - 2$. Setting $p(x) = 0$,we get $x - 2 = 0$,which implies $x = 2$. Here,$2$ is the zero of the polynomial,which is not $0$. Therefore,the zero of a polynomial can be any real number,not necessarily $0$.

(B) The given statement is False.
Justification: $A$ polynomial can have more than one zero. The number of zeroes of a polynomial depends on its degree. According to the Fundamental Theorem of Algebra,a polynomial of degree $n$ can have at most $n$ real zeroes. For example,a quadratic polynomial $(degree = 2)$ can have up to $2$ zeroes,and a cubic polynomial $(degree = 3)$ can have up to $3$ zeroes.

(B) Let $p(x) = x^{3}+3x^{2}+3x+1$.
To find the remainder when $p(x)$ is divided by $5+2x$,we use the Remainder Theorem.
First,find the zero of the divisor $5+2x$:
$5+2x = 0 \implies 2x = -5 \implies x = -\frac{5}{2}$.
Now,evaluate $p(-\frac{5}{2})$:
$p(-\frac{5}{2}) = (-\frac{5}{2})^{3} + 3(-\frac{5}{2})^{2} + 3(-\frac{5}{2}) + 1$
$= -\frac{125}{8} + 3(\frac{25}{4}) - \frac{15}{2} + 1$
$= -\frac{125}{8} + \frac{75}{4} - \frac{15}{2} + 1$
$= \frac{-125 + 150 - 60 + 8}{8}$
$= \frac{-27}{8}$.
Thus,the required remainder is $-\frac{27}{8}$.

(C) Let the polynomial be $p(x) = 3x^{3} - 4x^{2} + 7x - 5$.
To find the value at $x = -3$,we substitute $-3$ for $x$ in the polynomial:
$p(-3) = 3(-3)^{3} - 4(-3)^{2} + 7(-3) - 5$
Calculate the powers:
$p(-3) = 3(-27) - 4(9) - 21 - 5$
Perform the multiplications:
$p(-3) = -81 - 36 - 21 - 5$
Sum the negative values:
$p(-3) = -143$.

(B) zero of a polynomial $p(x)$ is a number $c$ such that $p(c) = 0$.
Let $p(x) = x - 3$.
To check if $-3$ is a zero,we calculate $p(-3)$:
$p(-3) = -3 - 3 = -6$.
Since $p(-3) = -6 \neq 0$,$-3$ is not a zero of the polynomial $x - 3$.
Therefore,the statement is False.

(A) zero of a polynomial $p(x)$ is a value $c$ such that $p(c) = 0$.
Let $p(x) = 3x + 1$.
To verify,substitute $x = -\frac{1}{3}$ into the polynomial:
$p\left(-\frac{1}{3}\right) = 3\left(-\frac{1}{3}\right) + 1$
$p\left(-\frac{1}{3}\right) = -1 + 1 = 0$.
Since $p\left(-\frac{1}{3}\right) = 0$,it is confirmed that $-\frac{1}{3}$ is a zero of the polynomial $3x + 1$. Therefore,the statement is True.

(FALSE) zero of a polynomial $p(y)$ is a value $c$ such that $p(c) = 0$.
Let $p(y) = 4 - 5y$.
Substitute $y = -\frac{4}{5}$ into the polynomial:
$p\left(-\frac{4}{5}\right) = 4 - 5\left(-\frac{4}{5}\right)$
$p\left(-\frac{4}{5}\right) = 4 + 4 = 8$.
Since $p\left(-\frac{4}{5}\right) \neq 0$,the statement is False.
Therefore,$-\frac{4}{5}$ is not a zero of $4 - 5y$.

(TRUE) zero of a polynomial $p(t)$ is a value $c$ such that $p(c) = 0$.
Let $p(t) = t^{2} - 2t$.
For $t = 0$:
$p(0) = (0)^{2} - 2(0) = 0 - 0 = 0$.
For $t = 2$:
$p(2) = (2)^{2} - 2(2) = 4 - 4 = 0$.
Since $p(0) = 0$ and $p(2) = 0$,the statement is True. Thus,$0$ and $2$ are the zeroes of the polynomial $t^{2} - 2t$.

(TRUE) zero of a polynomial $p(y)$ is a value $c$ such that $p(c) = 0$.
Let $p(y) = y^{2} + y - 6$.
Substitute $y = -3$ into the polynomial:
$p(-3) = (-3)^{2} + (-3) - 6$
$p(-3) = 9 - 3 - 6$
$p(-3) = 0$.
Since $p(-3) = 0$,it is True that $-3$ is a zero of the polynomial $y^{2} + y - 6$.

(D) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = x - 4$.
Setting the polynomial equal to zero,we get:
$x - 4 = 0$
Adding $4$ to both sides of the equation:
$x = 4$
Therefore,the zero of the polynomial $p(x) = x - 4$ is $4$.

(A) To find the zero of the polynomial $g(x) = 3 - 6x$,we set $g(x) = 0$.
$3 - 6x = 0$
Subtract $3$ from both sides:
$-6x = -3$
Divide both sides by $-6$:
$x = \frac{-3}{-6}$
Simplifying the fraction,we get:
$x = \frac{1}{2}$
Therefore,the zero of the polynomial $g(x) = 3 - 6x$ is $\frac{1}{2}$.

(B) To find the zero of the polynomial $q(x) = 2x - 7$,we set the polynomial equal to zero:
$q(x) = 0$
$2x - 7 = 0$
Adding $7$ to both sides,we get:
$2x = 7$
Dividing both sides by $2$,we get:
$x = \frac{7}{2}$
Therefore,the zero of the polynomial $q(x) = 2x - 7$ is $\frac{7}{2}$.

(C) To find the zero of the polynomial $h(y) = 2y$,we set the polynomial equal to zero.
$h(y) = 0$
$2y = 0$
Dividing both sides by $2$,we get:
$y = 0$
Therefore,$0$ is the zero of the polynomial $h(y) = 2y$.

(D) To find the zeroes of the polynomial $p(x)$,we set $p(x) = 0$.
Given $p(x) = (x-2)^{2} - (x+2)^{2}$.
Using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$,where $a = (x-2)$ and $b = (x+2)$:
$p(x) = [(x-2) - (x+2)] \cdot [(x-2) + (x+2)] = 0$
$p(x) = (x - 2 - x - 2) \cdot (x - 2 + x + 2) = 0$
$p(x) = (-4) \cdot (2x) = 0$
$-8x = 0$
$x = 0$
Thus,the zero of the polynomial is $0$.

(N/A) Let $p(x) = x^{2}-x-6$.
For $x = 3$:
$p(3) = (3)^{2} - 3 - 6 = 9 - 3 - 6 = 0$.
Since $p(3) = 0$,$3$ is a zero of the polynomial $x^{2}-x-6$.
For $x = 5$:
$p(5) = (5)^{2} - 5 - 6 = 25 - 5 - 6 = 14$.
Since $p(5) \neq 0$,$5$ is not a zero of the polynomial $x^{2}-x-6$.

(B) Given polynomial is $p(x)=5 x-8$.
To find the zero of the polynomial,we set $p(x)=0$.
So,$5 x-8=0$.
Adding $8$ to both sides,we get $5 x=8$.
Dividing both sides by $5$,we get $x=\frac{8}{5}$.
Therefore,the zero of the polynomial $p(x)=5 x-8$ is $\frac{8}{5}$.

(C) To find the value of the polynomial $p(x) = x^{2}-7x+12$ at $x=1$,we substitute $1$ for $x$ in the expression.
$p(1) = (1)^{2} - 7(1) + 12$
$p(1) = 1 - 7 + 12$
$p(1) = -6 + 12$
$p(1) = 6$
Therefore,the value of the polynomial at $x=1$ is $6$.

(A) To find the value of the polynomial $p(x) = x^{2} - 7x + 12$ at $x = 3$,we substitute $3$ for $x$ in the expression:
$p(3) = (3)^{2} - 7(3) + 12$
$p(3) = 9 - 21 + 12$
$p(3) = 21 - 21$
$p(3) = 0$
Therefore,the value of the polynomial at $x = 3$ is $0$.

(B) To find the value of the polynomial $p(x) = x^{2}-7x+12$ at $x=4$,we substitute $4$ for $x$ in the expression.
$p(4) = (4)^{2} - 7(4) + 12$
$p(4) = 16 - 28 + 12$
$p(4) = 28 - 28$
$p(4) = 0$
Therefore,the value of the polynomial at $x=4$ is $0$.

To find the values,we substitute $x = 1$,$x = 2$,and $x = 4$ into the polynomial $p(x) = x^{3} - 7x^{2} + 14x - 8$.
$1$. For $p(1)$:
$p(1) = (1)^{3} - 7(1)^{2} + 14(1) - 8 = 1 - 7 + 14 - 8 = 0$.
$2$. For $p(2)$:
$p(2) = (2)^{3} - 7(2)^{2} + 14(2) - 8 = 8 - 7(4) + 28 - 8 = 8 - 28 + 28 - 8 = 0$.
$3$. For $p(4)$:
$p(4) = (4)^{3} - 7(4)^{2} + 14(4) - 8 = 64 - 7(16) + 56 - 8 = 64 - 112 + 56 - 8 = 0$.
Thus,$p(1) = 0$,$p(2) = 0$,and $p(4) = 0$.

(N/A) To find the values of the polynomial $p(x) = x^{2} - 3x + 2$ at $x = 1, 2, 4$,we substitute the values into the expression:
$1$. For $x = 1$: $p(1) = (1)^{2} - 3(1) + 2 = 1 - 3 + 2 = 0$.
$2$. For $x = 2$: $p(2) = (2)^{2} - 3(2) + 2 = 4 - 6 + 2 = 0$.
$3$. For $x = 4$: $p(4) = (4)^{2} - 3(4) + 2 = 16 - 12 + 2 = 6$.
Thus,the values are $p(1) = 0$,$p(2) = 0$,and $p(4) = 6$.

(D) Let the polynomial be $p(x) = x^{2} - 2x - 15$.
To verify if $2$ is a zero,we calculate $p(2)$:
$p(2) = (2)^{2} - 2(2) - 15 = 4 - 4 - 15 = -15$.
Since $p(2) \neq 0$,$2$ is not a zero of the polynomial.
To verify if $5$ is a zero,we calculate $p(5)$:
$p(5) = (5)^{2} - 2(5) - 15 = 25 - 10 - 15 = 0$.
Since $p(5) = 0$,$5$ is a zero of the polynomial.

(C) To verify if a value is a zero of the polynomial $p(x) = x^{2} - 5x - 14$,we substitute the value into the polynomial and check if the result is $0$.
For $x = 3$:
$p(3) = (3)^{2} - 5(3) - 14$
$p(3) = 9 - 15 - 14$
$p(3) = -20$
Since $p(3) \neq 0$,$3$ is not a zero of the polynomial.
For $x = 7$:
$p(7) = (7)^{2} - 5(7) - 14$
$p(7) = 49 - 35 - 14$
$p(7) = 49 - 49$
$p(7) = 0$
Since $p(7) = 0$,$7$ is a zero of the polynomial.

(A) To find the zero of the polynomial $p(x) = 3x - 4$,we set $p(x) = 0$.
$3x - 4 = 0$
Adding $4$ to both sides,we get:
$3x = 4$
Dividing both sides by $3$,we get:
$x = \frac{4}{3}$
Therefore,the zero of the polynomial is $\frac{4}{3}$.

(D) To find the zero of the polynomial $p(t)$,we set $p(t) = 0$.
Given $p(t) = 7t - 21$.
Setting the polynomial equal to zero:
$7t - 21 = 0$
Add $21$ to both sides:
$7t = 21$
Divide both sides by $7$:
$t = \frac{21}{7}$
$t = 3$
Therefore,the zero of the polynomial is $3$.

(N/A) To find the zero of the polynomial $q(y)$,we set $q(y) = 0$.
So,$\pi y + 3.14 = 0$.
Subtracting $3.14$ from both sides,we get $\pi y = -3.14$.
Dividing both sides by $\pi$,we get $y = -\frac{3.14}{\pi}$.
Thus,the zero of the polynomial is $-\frac{3.14}{\pi}$.

(D) To find the zero of the polynomial $p(x)$,we set $p(x) = 0$.
$\frac{2}{3}x + \frac{5}{4} = 0$
Subtract $\frac{5}{4}$ from both sides:
$\frac{2}{3}x = -\frac{5}{4}$
Multiply both sides by $\frac{3}{2}$ to solve for $x$:
$x = -\frac{5}{4} \times \frac{3}{2}$
$x = -\frac{15}{8}$

(C) To find the zero of the polynomial $q(m)$,we set $q(m) = 0$.
$0.3m - 0.15 = 0$
$0.3m = 0.15$
$m = \frac{0.15}{0.3}$
$m = \frac{15}{30}$
$m = 0.5$
Therefore,the zero of the polynomial is $0.5$.