Mix Examples - Polynomials Questions in English

(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 $0.5$.

(B) Given the polynomial $p(x) = x^{2} - 4x + 3$.
First,calculate $p(2)$:
$p(2) = (2)^{2} - 4(2) + 3 = 4 - 8 + 3 = -1$.
Next,calculate $p(-1)$:
$p(-1) = (-1)^{2} - 4(-1) + 3 = 1 + 4 + 3 = 8$.
Then,calculate $p(1/2)$:
$p(1/2) = (1/2)^{2} - 4(1/2) + 3 = 1/4 - 2 + 3 = 1/4 + 1 = 5/4$.
Finally,substitute these values into the expression $p(2) - p(-1) + p(1/2)$:
$-1 - 8 + 5/4 = -9 + 5/4 = (-36 + 5) / 4 = -31/4$.

(N/A) $p(x)=21+10 x+x^{2}$
$\therefore p(x)=x^{2}+10 x+21$ and $g(x)=2+x$
$\therefore g(x)=x+2$
$\therefore$ Quotient $=x+8$ and remainder $=5$
Steps of division process:
Step $1:$ We write the dividend $21+10 x+x^{2}$ and the divisor $2+x$ in the standard form,i.e.,after arranging the terms in the descending order of their degrees. So,the dividend is $x^{2}+10 x+21$ and the divisor is $x+2$.
Step $2:$ We divide the first term of the dividend by the first term of the divisor,i.e.,we divide $x^{2}$ by $x$ and get $x$. This gives us the first term of the quotient.
$\frac{x^{2}}{x}=x=$ first term of quotient.
Step $3:$ We multiply the divisor by the first term of the quotient and subtract this product from the dividend,i.e.,we multiply $x+2$ by $x$ and subtract the product $x^{2}+2 x$ from the dividend $x^{2}+10 x+21$. This gives us the remainder as $8 x+21$.
Step $4:$ We treat the remainder $8 x+21$ as the new dividend. The divisor remains the same. We repeat step $2$ to get the next term of the quotient,i.e.,we divide the first term $8 x$ of the (new) dividend by the first term $x$ of the divisor and obtain $8$.
Thus,$8$ is the second term in the quotient.
$\frac{8 x}{x}=8=$ second term of quotient.
New quotient $=x+8$
Step $5:$ We multiply the divisor by the second term of the quotient and subtract the product from the dividend. That is,we multiply $x+2$ by $8$ and subtract the product $8 x+16$ from the dividend $8 x+21$. This gives us $5$ as the remainder.
This process continues till the remainder is $0$ or the degree of the new dividend is less than the degree of the divisor. At this stage,this new dividend becomes the remainder and the sum of the quotients gives us the whole quotient.
Step $6:$ Thus,the quotient in full is $x+8$ and the remainder is $5.$

(N/A) To divide $p(x) = x^{3} + 7x^{2} + 14x + 1$ by $x + 3$,we use polynomial long division:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x)$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x + 3)$ to get $x^{3} + 3x^{2}$. Subtract this from the dividend to get $4x^{2} + 14x + 1$.
$3$. Divide $4x^{2}$ by $x$ to get $4x$. Multiply $4x$ by $(x + 3)$ to get $4x^{2} + 12x$. Subtract this to get $2x + 1$.
$4$. Divide $2x$ by $x$ to get $2$. Multiply $2$ by $(x + 3)$ to get $2x + 6$. Subtract this to get $-5$.
Thus,the quotient is $x^{2} + 4x + 2$ and the remainder is $-5$.

(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 $5$.

(A) Let $p(x) = x^{3} + 9x^{2} + 26x + 24$ be the given polynomial.
To check if $(x+2)$ is a factor,we find the zero of the linear polynomial $x+2$ by setting $x+2 = 0$,which gives $x = -2$.
According to the factor theorem,if $p(-2) = 0$,then $(x+2)$ is a factor of $p(x)$.
Now,substitute $x = -2$ into $p(x)$:
$p(-2) = (-2)^{3} + 9(-2)^{2} + 26(-2) + 24$
$p(-2) = -8 + 9(4) - 52 + 24$
$p(-2) = -8 + 36 - 52 + 24$
$p(-2) = 60 - 60 = 0$
Since $p(-2) = 0$,by the factor theorem,$(x+2)$ is a factor of $x^{3} + 9x^{2} + 26x + 24$.

(A) The polynomial $p(x)$ will be a multiple of $x + 2$ if and only if $x + 2$ divides $p(x)$ leaving a remainder of $0$.
According to the Factor Theorem,if $p(a) = 0$,then $(x - a)$ is a factor of $p(x)$.
Setting $x + 2 = 0$,we get $x = -2$.
Now,we evaluate $p(-2)$:
$p(-2) = (-2)^{3} + 9(-2)^{2} + 26(-2) + 24$
$p(-2) = -8 + 9(4) - 52 + 24$
$p(-2) = -8 + 36 - 52 + 24$
$p(-2) = (-8 - 52) + (36 + 24)$
$p(-2) = -60 + 60$
$p(-2) = 0$
Since the remainder is $0$,$(x + 2)$ is a factor of $p(x)$.
Therefore,$p(x)$ is a multiple of $x + 2$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $2x^2 - 7x - 15$ by $x + 1$.
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$: $2x^2 / x = 2x$. This is the first term of the quotient.
$2$. Multiply the divisor $(x + 1)$ by $2x$: $2x(x + 1) = 2x^2 + 2x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + 2x) = -9x - 15$.
$4$. Divide the first term of the new expression $(-9x)$ by the first term of the divisor $(x)$: $-9x / x = -9$. This is the second term of the quotient.
$5$. Multiply the divisor $(x + 1)$ by $-9$: $-9(x + 1) = -9x - 9$.
$6$. Subtract this from the current expression: $(-9x - 15) - (-9x - 9) = -9x - 15 + 9x + 9 = -6$.
Thus,the quotient is $2x - 9$ and the remainder is $-6$.

(N/A) To find the quotient and remainder,we perform polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.
$2$. Multiply the divisor $(2x+1)$ by $x$: $x(2x+1) = 2x^2 + x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + x) = -8x - 15$.
$4$. Divide the first term of the new expression $(-8x)$ by the first term of the divisor $(2x)$: $-8x / 2x = -4$. This is the second term of the quotient.
$5$. Multiply the divisor $(2x+1)$ by $-4$: $-4(2x+1) = -8x - 4$.
$6$. Subtract this from the current expression: $(-8x - 15) - (-8x - 4) = -8x - 15 + 8x + 4 = -11$.
Therefore,the Quotient is $x-4$ and the Remainder is $-11$.

(N/A) To divide $2x^2 - 7x - 15$ by $2x + 3$,we use polynomial long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.
$2$. Multiply the divisor $(2x + 3)$ by $x$: $x(2x + 3) = 2x^2 + 3x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 + 3x) = -10x - 15$.
$4$. Divide the first term of the new expression $(-10x)$ by the first term of the divisor $(2x)$: $-10x / 2x = -5$. This is the second term of the quotient.
$5$. Multiply the divisor $(2x + 3)$ by $-5$: $-5(2x + 3) = -10x - 15$.
$6$. Subtract this from the current expression: $(-10x - 15) - (-10x - 15) = 0$.
Therefore,the quotient is $x - 5$ and the remainder is $0$.

(A) To divide $2x^2 - 7x - 15$ by $2x - 3$,we use long division:
$1$. Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(2x)$: $2x^2 / 2x = x$. This is the first term of the quotient.
$2$. Multiply the divisor $(2x - 3)$ by $x$: $x(2x - 3) = 2x^2 - 3x$.
$3$. Subtract this from the dividend: $(2x^2 - 7x - 15) - (2x^2 - 3x) = -4x - 15$.
$4$. Divide the first term of the new expression $(-4x)$ by the first term of the divisor $(2x)$: $-4x / 2x = -2$. This is the second term of the quotient.
$5$. Multiply the divisor $(2x - 3)$ by $-2$: $-2(2x - 3) = -4x + 6$.
$6$. Subtract this from the current expression: $(-4x - 15) - (-4x + 6) = -4x - 15 + 4x - 6 = -21$.
Thus,the quotient is $x - 2$ and the remainder is $-21$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $2x^2 - 7x - 15$ by $x - 2$.
Step $1$: Divide the first term of the dividend $(2x^2)$ by the first term of the divisor $(x)$ to get $2x$.
Step $2$: Multiply $2x$ by $(x - 2)$ to get $2x^2 - 4x$.
Step $3$: Subtract $(2x^2 - 4x)$ from $(2x^2 - 7x)$ to get $-3x$.
Step $4$: Bring down the next term $-15$ to get $-3x - 15$.
Step $5$: Divide the first term of the new expression $(-3x)$ by the first term of the divisor $(x)$ to get $-3$.
Step $6$: Multiply $-3$ by $(x - 2)$ to get $-3x + 6$.
Step $7$: Subtract $(-3x + 6)$ from $(-3x - 15)$ to get $-15 - 6 = -21$.
Thus,the Quotient $= 2x - 3$ and the Remainder $= -21$.

(A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x-1)$.
Step $1$: Divide the first term of the dividend $x^3$ by the first term of the divisor $x$ to get $x^2$.
Step $2$: Multiply $x^2$ by $(x-1)$ to get $(x^3-x^2)$. Subtract this from the dividend: $(x^3+x^2-10x+8) - (x^3-x^2) = 2x^2-10x+8$.
Step $3$: Divide the first term of the new polynomial $2x^2$ by $x$ to get $2x$.
Step $4$: Multiply $2x$ by $(x-1)$ to get $(2x^2-2x)$. Subtract this: $(2x^2-10x+8) - (2x^2-2x) = -8x+8$.
Step $5$: Divide the first term $-8x$ by $x$ to get $-8$.
Step $6$: Multiply $-8$ by $(x-1)$ to get $(-8x+8)$. Subtract this: $(-8x+8) - (-8x+8) = 0$.
Thus,the Quotient is $x^{2}+2x-8$ and the Remainder is $0$.

(A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x-2)$:
$1$. Divide the first term of the dividend $(x^{3})$ by the first term of the divisor $(x)$ to get $x^{2}$.
$2$. Multiply $x^{2}$ by $(x-2)$ to get $x^{3}-2x^{2}$.
$3$. Subtract $(x^{3}-2x^{2})$ from $(x^{3}+x^{2}-10x+8)$ to get $3x^{2}-10x+8$.
$4$. Divide $3x^{2}$ by $x$ to get $3x$. Multiply $3x$ by $(x-2)$ to get $3x^{2}-6x$.
$5$. Subtract $(3x^{2}-6x)$ from $(3x^{2}-10x+8)$ to get $-4x+8$.
$6$. Divide $-4x$ by $x$ to get $-4$. Multiply $-4$ by $(x-2)$ to get $-4x+8$.
$7$. Subtract $(-4x+8)$ from $(-4x+8)$ to get $0$.
Thus,the Quotient $= x^{2}+3x-4$ and the Remainder $= 0$.

(N/A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x+3)$:
$1$. Divide the first term of the dividend $(x^3)$ by the first term of the divisor $(x)$ to get $x^2$.
$2$. Multiply $x^2$ by $(x+3)$ to get $x^3+3x^2$. Subtract this from the dividend: $(x^3+x^2-10x+8) - (x^3+3x^2) = -2x^2-10x+8$.
$3$. Divide the new first term $(-2x^2)$ by $x$ to get $-2x$.
$4$. Multiply $-2x$ by $(x+3)$ to get $-2x^2-6x$. Subtract this: $(-2x^2-10x+8) - (-2x^2-6x) = -4x+8$.
$5$. Divide the new first term $(-4x)$ by $x$ to get $-4$.
$6$. Multiply $-4$ by $(x+3)$ to get $-4x-12$. Subtract this: $(-4x+8) - (-4x-12) = 20$.
Thus,the Quotient $= x^{2}-2x-4$ and the Remainder $= 20$.

(A) To find the quotient and remainder,we perform polynomial long division of $(x^{3}+x^{2}-10x+8)$ by $(x+4)$:
$1$. Divide the first term of the dividend $(x^3)$ by the first term of the divisor $(x)$ to get $x^2$.
$2$. Multiply $x^2$ by $(x+4)$ to get $x^3+4x^2$. Subtract this from the dividend: $(x^3+x^2-10x+8) - (x^3+4x^2) = -3x^2-10x+8$.
$3$. Divide the new first term $(-3x^2)$ by $x$ to get $-3x$.
$4$. Multiply $-3x$ by $(x+4)$ to get $-3x^2-12x$. Subtract this from the current remainder: $(-3x^2-10x+8) - (-3x^2-12x) = 2x+8$.
$5$. Divide the new first term $(2x)$ by $x$ to get $2$.
$6$. Multiply $2$ by $(x+4)$ to get $2x+8$. Subtract this: $(2x+8) - (2x+8) = 0$.
Thus,the Quotient $= x^{2}-3x+2$ and the Remainder $= 0$.

(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 $x + 1$,which can be written as $x - (-1)$.
Therefore,we need to find $p(-1)$.
Given $p(x) = x^3 + x^2 - 26x + 24$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^3 + (-1)^2 - 26(-1) + 24$
$p(-1) = -1 + 1 + 26 + 24$
$p(-1) = 0 + 50$
$p(-1) = 50$.
Thus,the remainder is $50$.

(D) According to the remainder theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,$p(x) = x^{3} + x^{2} - 26x + 24$ and the divisor is $(x - 1)$,so $a = 1$.
To find the remainder,we calculate $p(1)$:
$p(1) = (1)^{3} + (1)^{2} - 26(1) + 24$
$p(1) = 1 + 1 - 26 + 24$
$p(1) = 2 - 26 + 24$
$p(1) = -24 + 24 = 0$.
Therefore,the remainder is $0$.

(A) According to the remainder theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,the divisor is $x + 4$,which can be written as $x - (-4)$.
Therefore,$a = -4$.
Now,substitute $x = -4$ into the polynomial $p(x) = x^3 + x^2 - 26x + 24$:
$p(-4) = (-4)^3 + (-4)^2 - 26(-4) + 24$
$p(-4) = -64 + 16 + 104 + 24$
$p(-4) = -64 + 144$
$p(-4) = 80$.
Thus,the remainder is $80$.

(B) According to the Remainder Theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,$p(x) = x^{3} + x^{2} - 26x + 24$ and the divisor is $x - 4$,so $a = 4$.
To find the remainder,we calculate $p(4)$:
$p(4) = (4)^{3} + (4)^{2} - 26(4) + 24$
$p(4) = 64 + 16 - 104 + 24$
$p(4) = 80 - 104 + 24$
$p(4) = -24 + 24 = 0$.
Therefore,the remainder is $0$.

(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 $0$.

(D) According to the remainder theorem,if a polynomial $p(x)$ is divided by $(x - a)$,the remainder is $p(a)$.
Here,$p(x) = x^{3} + x^{2} - 26x + 24$ and the divisor is $x - 6$,so $a = 6$.
Now,substitute $x = 6$ into the polynomial:
$p(6) = (6)^{3} + (6)^{2} - 26(6) + 24$
$p(6) = 216 + 36 - 156 + 24$
$p(6) = 252 - 156 + 24$
$p(6) = 96 + 24$
$p(6) = 120$
Therefore,the remainder is $120$.

(A) Let $p(x) = x^{3} + 13x^{2} + ax + 35$.
Since $x+5$ is a factor of $p(x)$,by the Factor Theorem,$p(-5) = 0$.
Substituting $x = -5$ into the polynomial:
$(-5)^{3} + 13(-5)^{2} + a(-5) + 35 = 0$
$-125 + 13(25) - 5a + 35 = 0$
$-125 + 325 - 5a + 35 = 0$
$235 - 5a = 0$
$5a = 235$
$a = 47$.

(A) Let the polynomial be $p(x) = x^{3} + ax^{2} - x + 30$.
Since $(x+2)$ is a factor of $p(x)$,by the Factor Theorem,$p(-2) = 0$.
Substituting $x = -2$ into the polynomial:
$p(-2) = (-2)^{3} + a(-2)^{2} - (-2) + 30 = 0$
$-8 + 4a + 2 + 30 = 0$
$4a + 24 = 0$
$4a = -24$
$a = -6$.

(B) Let $p(x) = x^{3} + ax^{2} + 19x + 20$.
According to the Remainder Theorem,when $p(x)$ is divided by $(x + 3)$,the remainder is $p(-3)$.
Given that the remainder is $a$,we have $p(-3) = a$.
Substituting $x = -3$ into the polynomial:
$(-3)^{3} + a(-3)^{2} + 19(-3) + 20 = a$
$-27 + 9a - 57 + 20 = a$
$9a - 64 = a$
$8a = 64$
$a = 8$.

(A) According to the Remainder Theorem,if $p(x)$ is divided by $(x + 1)$,the remainder is $p(-1)$.
Given $p(-1) = 16$,we have:
$2(-1)^3 - 3(-1)^2 + a(-1) - 3a + 9 = 16$
$-2 - 3 - a - 3a + 9 = 16$
$4 - 4a = 16$
$-4a = 12$
$a = -3$.
Now,substitute $a = -3$ into $p(x)$:
$p(x) = 2x^3 - 3x^2 - 3x - 3(-3) + 9 = 2x^3 - 3x^2 - 3x + 18$.
To find the remainder when $p(x)$ is divided by $(x + 2)$,we calculate $p(-2)$:
$p(-2) = 2(-2)^3 - 3(-2)^2 - 3(-2) + 18$
$p(-2) = 2(-8) - 3(4) + 6 + 18$
$p(-2) = -16 - 12 + 6 + 18 = -4$.
Thus,$a = -3$ and the remainder is $-4$.

(A) According to the Remainder Theorem,if a polynomial $f(x)$ is divided by $(x - c)$,the remainder is $f(c)$.
For $p(x) = x^{3} + 2x^{2} - 5ax - 7$ divided by $(x + 1)$,$R_{1} = p(-1) = (-1)^{3} + 2(-1)^{2} - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6$.
For $q(x) = x^{3} + ax^{2} - 12x + 6$ divided by $(x - 2)$,$R_{2} = q(2) = (2)^{3} + a(2)^{2} - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10$.
Given $2R_{1} + R_{2} = 6$,substitute the expressions for $R_{1}$ and $R_{2}$:
$2(5a - 6) + (4a - 10) = 6$
$10a - 12 + 4a - 10 = 6$
$14a - 22 = 6$
$14a = 28$
$a = 2$.

(A) Let $p(x) = 2x^3 + 21x^2 + 67x + 60$.
To check if $2x + 3$ is a factor,we find its zero by setting $2x + 3 = 0$,which gives $x = -\frac{3}{2}$.
According to the factor theorem,if $p(-\frac{3}{2}) = 0$,then $2x + 3$ is a factor.
$p(-\frac{3}{2}) = 2(-\frac{3}{2})^3 + 21(-\frac{3}{2})^2 + 67(-\frac{3}{2}) + 60$
$= 2(-\frac{27}{8}) + 21(\frac{9}{4}) - \frac{201}{2} + 60$
$= -\frac{27}{4} + \frac{189}{4} - \frac{402}{4} + \frac{240}{4}$
$= \frac{-27 + 189 - 402 + 240}{4}$
$= \frac{429 - 429}{4} = 0$.
Since $p(-\frac{3}{2}) = 0$,$2x + 3$ is a factor of the given polynomial.

(N/A) Let $p(x) = x^{2}-7x+12$.
According to the factor theorem,if $(x-a)$ is a factor of $p(x)$,then $p(a) = 0$.
We look for factors of the constant term $12$. The possible factors are $\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12$.
Testing $x = 3$:
$p(3) = (3)^{2} - 7(3) + 12 = 9 - 21 + 12 = 0$.
Since $p(3) = 0$,$(x-3)$ is a factor of $p(x)$.
Testing $x = 4$:
$p(4) = (4)^{2} - 7(4) + 12 = 16 - 28 + 12 = 0$.
Since $p(4) = 0$,$(x-4)$ is a factor of $p(x)$.
Therefore,the factors of $x^{2}-7x+12$ are $(x-3)(x-4)$.

(A) To factorise $10 x^{2}-x-24$ by splitting the middle term,we need to find two numbers $p$ and $q$ such that $p+q = -1$ (the coefficient of $x$) and $p \times q = 10 \times (-24) = -240$ (the product of the coefficient of $x^{2}$ and the constant term).
We look for factors of $-240$ that add up to $-1$. These numbers are $15$ and $-16$,because $15 + (-16) = -1$ and $15 \times (-16) = -240$.
Now,rewrite the middle term $-x$ as $15x - 16x$:
$10 x^{2} - x - 24 = 10 x^{2} + 15 x - 16 x - 24$
Group the terms to factor out common factors:
$= (10 x^{2} + 15 x) - (16 x + 24)$
$= 5 x(2 x + 3) - 8(2 x + 3)$
Finally,factor out the common binomial $(2 x + 3)$:
$= (2 x + 3)(5 x - 8)$

(A) Let $p(x) = x^{3}+x^{2}-26 x+24$.
Sum of all the coefficients of the polynomial is $1+1-26+24 = 0$.
Since the sum of coefficients is $0$,$(x-1)$ is a factor of $p(x)$.
Now,we rewrite the polynomial to factorize it:
$x^{3}+x^{2}-26 x+24 = x^{3}-x^{2}+2x^{2}-2x-24x+24$
$= x^{2}(x-1)+2x(x-1)-24(x-1)$
$= (x-1)(x^{2}+2x-24)$
Now,factorize the quadratic expression $(x^{2}+2x-24)$ by splitting the middle term:
$x^{2}+6x-4x-24 = x(x+6)-4(x+6) = (x-4)(x+6)$
Therefore,the factors are $(x-1)(x-4)(x+6)$.

(D) Let $p(x) = x^{3}-x^{2}-17 x-15$.
To find a factor,we test values using the Factor Theorem. Let us check $x = -1$:
$p(-1) = (-1)^{3} - (-1)^{2} - 17(-1) - 15 = -1 - 1 + 17 - 15 = 0$.
Since $p(-1) = 0$,$(x+1)$ is a factor of $p(x)$.
Now,we divide $p(x)$ by $(x+1)$ or split the terms:
$x^{3}-x^{2}-17 x-15 = x^{3}+x^{2}-2x^{2}-2x-15x-15$
$= x^{2}(x+1) - 2x(x+1) - 15(x+1)$
$= (x+1)(x^{2}-2x-15)$
Now,factorize the quadratic expression $(x^{2}-2x-15)$ by splitting the middle term:
$x^{2}-5x+3x-15 = x(x-5)+3(x-5) = (x-5)(x+3)$.
Thus,the factors are $(x+1)(x-5)(x+3)$.

(A) According to the Factor Theorem,$(x-1)$ is a factor of $p(x)$ if $p(1) = 0$.
Substitute $x = 1$ into the polynomial:
$p(1) = (1)^{3} - 7(1)^{2} + 14(1) - 8$
$p(1) = 1 - 7 + 14 - 8$
$p(1) = 15 - 15 = 0$
Since $p(1) = 0$,$(x-1)$ is a factor of the given polynomial.

(A) To determine if $(x-1)$ is a factor of the polynomial $p(x) = x^{3}+4x^{2}+x-6$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$a = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = (1)^{3} + 4(1)^{2} + (1) - 6$
$p(1) = 1 + 4(1) + 1 - 6$
$p(1) = 1 + 4 + 1 - 6$
$p(1) = 6 - 6 = 0$
Since $p(1) = 0$,$(x-1)$ is a factor of the given polynomial.

(B) To determine if $(x-1)$ is a factor of the polynomial $p(x) = x^{3}+6x^{2}-9x-14$,we use the Factor Theorem.
According to the Factor Theorem,$(x-c)$ is a factor of $p(x)$ if $p(c) = 0$.
Here,$c = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = (1)^{3} + 6(1)^{2} - 9(1) - 14$
$p(1) = 1 + 6 - 9 - 14$
$p(1) = 7 - 23$
$p(1) = -16$
Since $p(1) \neq 0$,$(x-1)$ is not a factor of the given polynomial.

(A) To determine if $(x-1)$ is a factor of the polynomial $p(x) = 2x^3 + 5x^2 - x - 6$,we use the Factor Theorem.
According to the Factor Theorem,$(x-a)$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$a = 1$.
Substitute $x = 1$ into the polynomial:
$p(1) = 2(1)^3 + 5(1)^2 - (1) - 6$
$p(1) = 2(1) + 5(1) - 1 - 6$
$p(1) = 2 + 5 - 1 - 6$
$p(1) = 7 - 7 = 0$.
Since $p(1) = 0$,$(x-1)$ is a factor of the given polynomial.

(A) To determine if $(x+1)$ is a factor of the polynomial $p(x) = x^{3} + 10x^{2} + 23x + 14$,we use the Factor Theorem.
According to the Factor Theorem,$(x+1)$ is a factor of $p(x)$ if $p(-1) = 0$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} + 10(-1)^{2} + 23(-1) + 14$
$p(-1) = -1 + 10(1) - 23 + 14$
$p(-1) = -1 + 10 - 23 + 14$
$p(-1) = 9 - 23 + 14$
$p(-1) = -14 + 14 = 0$
Since $p(-1) = 0$,$(x+1)$ is indeed a factor of the given polynomial.

(A) To check if $(x+1)$ is a factor of $p(x) = x^{3} - 5x^{2} + 2x + 8$,we use the Factor Theorem.
According to the Factor Theorem,$(x+1)$ is a factor of $p(x)$ if $p(-1) = 0$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} - 5(-1)^{2} + 2(-1) + 8$
$p(-1) = -1 - 5(1) - 2 + 8$
$p(-1) = -1 - 5 - 2 + 8$
$p(-1) = -8 + 8 = 0$
Since $p(-1) = 0$,$(x+1)$ is a factor of the given polynomial.

(B) To check if $(x+1)$ is a factor of $p(x) = x^{3} - 2x^{2} - 5x + 6$,we use the Factor Theorem.
According to the Factor Theorem,$(x+1)$ is a factor if $p(-1) = 0$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} - 2(-1)^{2} - 5(-1) + 6$
$p(-1) = -1 - 2(1) + 5 + 6$
$p(-1) = -1 - 2 + 5 + 6$
$p(-1) = 8$
Since $p(-1) \neq 0$,$(x+1)$ is not a factor of the given polynomial.

(A) To determine if $(x+1)$ is a factor of $P(x) = 6x^3 + 11x^2 - 5x - 12$,we use the Factor Theorem.
According to the Factor Theorem,$(x+1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$.
Substitute $x = -1$ into the polynomial:
$P(-1) = 6(-1)^3 + 11(-1)^2 - 5(-1) - 12$
$P(-1) = 6(-1) + 11(1) + 5 - 12$
$P(-1) = -6 + 11 + 5 - 12$
$P(-1) = 5 + 5 - 12$
$P(-1) = 10 - 12$
$P(-1) = -2$
Since $P(-1) \neq 0$,$(x+1)$ is not a factor of the given polynomial.

(A) To factorise the quadratic polynomial $x^{2}+10x+16$ by splitting the middle term,we need to find two numbers whose sum is $10$ (the coefficient of $x$) and whose product is $16$ (the constant term).
$1$. Identify the factors of $16$: $(1, 16), (2, 8), (4, 4)$.
$2$. Among these,the pair $(2, 8)$ adds up to $10$.
$3$. Rewrite the middle term $10x$ as $2x + 8x$:
$x^{2} + 2x + 8x + 16$
$4$. Group the terms and factor out the common elements:
$x(x + 2) + 8(x + 2)$
$5$. Factor out the common binomial $(x + 2)$:
$(x + 2)(x + 8)$

(A) To factorise the quadratic polynomial $x^{2}-12x+20$,we need to find two numbers whose product is $20$ and whose sum is $-12$.
These two numbers are $-2$ and $-10$,since $(-2) \times (-10) = 20$ and $(-2) + (-10) = -12$.
Now,split the middle term $-12x$ as $-2x - 10x$:
$x^{2} - 2x - 10x + 20$
Group the terms:
$(x^{2} - 2x) - (10x - 20)$
Factor out the common terms from each group:
$x(x - 2) - 10(x - 2)$
Finally,factor out the common binomial $(x - 2)$:
$(x - 2)(x - 10)$

(A) To factorise $6x^2 + 19x + 10$ by splitting the middle term,we need to find two numbers whose product is $6 \times 10 = 60$ and whose sum is $19$.
These two numbers are $15$ and $4$,since $15 \times 4 = 60$ and $15 + 4 = 19$.
Now,rewrite the middle term $19x$ as $15x + 4x$:
$6x^2 + 15x + 4x + 10$
Group the terms to factor out common factors:
$(6x^2 + 15x) + (4x + 10)$
$3x(2x + 5) + 2(2x + 5)$
Finally,factor out the common binomial $(2x + 5)$:
$(2x + 5)(3x + 2)$

(N/A) To factorise the quadratic polynomial $x^{2}+14x+33$ by splitting the middle term,we need to find two numbers whose sum is $14$ and whose product is $33$.
$1$. Identify the coefficients: $a=1, b=14, c=33$.
$2$. Find two numbers $p$ and $q$ such that $p+q=14$ and $p \times q=33$.
$3$. The factors of $33$ are $(1, 33)$ and $(3, 11)$.
$4$. Since $3+11=14$,the numbers are $3$ and $11$.
$5$. Rewrite the middle term: $x^{2}+3x+11x+33$.
$6$. Factor by grouping: $x(x+3)+11(x+3)$.
$7$. Final factored form: $(x+3)(x+11)$.