Zeroes of a Polynomial Questions in English

(C) Let the polynomial be $p(x) = x^{3} + 7x^{2} + 17x + 25$.
According to the remainder theorem,when $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 4$,which can be written as $x - (-4)$.
Therefore,we set $x + 4 = 0$,which gives $x = -4$.
Now,we calculate $p(-4)$:
$p(-4) = (-4)^{3} + 7(-4)^{2} + 17(-4) + 25$
$p(-4) = -64 + 7(16) - 68 + 25$
$p(-4) = -64 + 112 - 68 + 25$
$p(-4) = 137 - 132$
$p(-4) = 5$
Thus,the remainder is $5$.

(C) According to the remainder theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 6$,which can be written as $x - (-6)$.
Therefore,we need to find $p(-6)$.
Given $p(x) = x^{3} + x^{2} - 26x + 24$.
Substituting $x = -6$ into the polynomial:
$p(-6) = (-6)^{3} + (-6)^{2} - 26(-6) + 24$
$p(-6) = -216 + 36 + 156 + 24$
$p(-6) = -216 + 216$
$p(-6) = 0$.
Thus,the remainder is $0$.

(C) Let the polynomial be $p(x) = x^{2} + 9x + 20$.
According to the Remainder Theorem,if $p(x)$ is divided by $(x + 2)$,the remainder is $p(-2)$.
Substitute $x = -2$ into the polynomial:
$p(-2) = (-2)^{2} + 9(-2) + 20$
$p(-2) = 4 - 18 + 20$
$p(-2) = 6$.
For the polynomial to be divisible by $(x + 2)$,the remainder must be $0$.
Therefore,we must subtract the remainder $6$ from the polynomial $p(x)$ so that the new polynomial becomes divisible by $(x + 2)$.

(B) To find the zero of the polynomial $p(x) = 2x + 3$,we set $p(x) = 0$.
$2x + 3 = 0$
$2x = -3$
$x = -\frac{3}{2}$
Since the calculated zero is $-\frac{3}{2}$ and not $\frac{3}{2}$,the given statement is False.

(B) To find $p(-2)$,substitute $x = -2$ into the polynomial $p(x) = x^{3} + 9x^{2} + 26x + 24$.
$p(-2) = (-2)^{3} + 9(-2)^{2} + 26(-2) + 24$
$p(-2) = -8 + 9(4) - 52 + 24$
$p(-2) = -8 + 36 - 52 + 24$
$p(-2) = (-8 - 52) + (36 + 24)$
$p(-2) = -60 + 60$
$p(-2) = 0$

(D) To find the zeros of the polynomial $p(x) = x^{3} + 7x^{2} + 11x + 5$,we test the given options by substituting them into the polynomial.
If $x = -1$:
$p(-1) = (-1)^{3} + 7(-1)^{2} + 11(-1) + 5$
$p(-1) = -1 + 7(1) - 11 + 5$
$p(-1) = -1 + 7 - 11 + 5$
$p(-1) = 12 - 12 = 0$
Since $p(-1) = 0$,$x = -1$ is a zero of the polynomial $p(x)$.

(C) To find the zeros of the polynomial $p(x) = x^{3}-3x^{2}-5x+15$,we set $p(x) = 0$.
Factor by grouping:
$x^{2}(x-3) - 5(x-3) = 0$
$(x^{2}-5)(x-3) = 0$
The zeros are found by setting each factor to zero:
$x-3 = 0 \implies x = 3$
$x^{2}-5 = 0 \implies x^{2} = 5 \implies x = \pm\sqrt{5}$
Comparing these values with the given options,$3$ is one of the zeros.

(B) To find the zero of the polynomial $p(x) = 5x - 10$,we set $p(x) = 0$.
$5x - 10 = 0$
$5x = 10$
$x = \frac{10}{5}$
$x = 2$
Therefore,the zero of the polynomial is $2$.

(A) To find the zero of the polynomial $p(x) = bx + m$,we set $p(x) = 0$.
$bx + m = 0$
$bx = -m$
$x = -\frac{m}{b}$
Therefore,the zero of the polynomial is $-\frac{m}{b}$.

(A) According to the Factor Theorem,if $p(a) = 0$ for a polynomial $p(x)$,then $(x - a)$ is a factor of $p(x)$.
Given that $p(3) = 0$,we substitute $a = 3$ into the theorem.
Therefore,$(x - 3)$ is a factor of $p(x)$.