Factorisation of Polynomials Questions in English

(B) Let $p(x) = 8x^4 + 4x^3 - 16x^2 + 10x + m$.
Since $(2x - 1)$ is a factor of $p(x)$,by the factor theorem,$p(\frac{1}{2}) = 0$.
Substituting $x = \frac{1}{2}$ into the polynomial:
$8(\frac{1}{2})^4 + 4(\frac{1}{2})^3 - 16(\frac{1}{2})^2 + 10(\frac{1}{2}) + m = 0$
$8(\frac{1}{16}) + 4(\frac{1}{8}) - 16(\frac{1}{4}) + 5 + m = 0$
$\frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$
$1 - 4 + 5 + m = 0$
$2 + m = 0$
Therefore,$m = -2$.

(C) Let $p(x) = ax^3 + x^2 - 2x + 4a - 9$.
According to the factor theorem,if $(x+1)$ is a factor of $p(x)$,then $p(-1) = 0$.
Substituting $x = -1$ into the polynomial:
$a(-1)^3 + (-1)^2 - 2(-1) + 4a - 9 = 0$
Simplifying the expression:
$-a + 1 + 2 + 4a - 9 = 0$
Combining like terms:
$3a - 6 = 0$
Solving for $a$:
$3a = 6$
$a = 2$

(A) To factorise the quadratic expression $x^{2}+9x+18$,we need to find two numbers $p$ and $q$ such that their sum $p+q = 9$ and their product $pq = 18$.
By observing the factors of $18$,we find that $6+3 = 9$ and $6 \times 3 = 18$.
Now,we split the middle term $9x$ as $6x + 3x$:
$x^{2}+9x+18 = x^{2}+6x+3x+18$
Next,we group the terms and factor out the common terms:
$= x(x+6) + 3(x+6)$
Finally,taking $(x+6)$ as a common factor:
$= (x+6)(x+3)$

(N/A) To factorise the quadratic expression $6x^{2} + 7x - 3$,we use the splitting the middle term method.
We need to find two numbers $p$ and $q$ such that $p + q = 7$ (coefficient of $x$) and $p \times q = 6 \times (-3) = -18$ (product of coefficient of $x^{2}$ and constant term).
Clearly,$9 + (-2) = 7$ and $9 \times (-2) = -18$.
Now,rewrite the middle term $7x$ as $9x - 2x$:
$6x^{2} + 9x - 2x - 3$
Group the terms to factor out common factors:
$= 3x(2x + 3) - 1(2x + 3)$
Finally,factor out the common binomial $(2x + 3)$:
$= (2x + 3)(3x - 1)$

(A) To factorise the quadratic polynomial $2 x^{2}-7 x-15$,we look for two numbers $p$ and $q$ such that their sum is the coefficient of $x$,which is $-7$,and their product is the product of the coefficient of $x^{2}$ and the constant term,which is $2 \times (-15) = -30$.
We need $p+q = -7$ and $p q = -30$.
By testing factors of $-30$,we find that $-10$ and $3$ satisfy these conditions: $(-10) + 3 = -7$ and $(-10) \times 3 = -30$.
Now,split the middle term $-7 x$ as $-10 x + 3 x$:
$2 x^{2}-7 x-15 = 2 x^{2}-10 x+3 x-15$
Factor by grouping:
$= 2 x(x-5) + 3(x-5)$
Taking $(x-5)$ as a common factor:
$= (x-5)(2 x+3)$

(B) To factorise $84 - 2r - 2r^2$,first rearrange the terms in descending order of powers of $r$:
$-2r^2 - 2r + 84$
Factor out the common term $-2$:
$-2(r^2 + r - 42)$
Now,factor the quadratic expression $r^2 + r - 42$. We need two numbers whose product is $-42$ and whose sum is $1$:
These numbers are $7$ and $-6$.
So,$r^2 + 7r - 6r - 42 = r(r + 7) - 6(r + 7) = (r - 6)(r + 7)$.
Combining these,the final factorised form is:
$-2(r - 6)(r + 7)$

(D) Let $f(x) = 2x^{3}-3x^{2}-17x+30$ be the given polynomial.
By testing values,we find $f(2) = 2(8) - 3(4) - 17(2) + 30 = 16 - 12 - 34 + 30 = 0$. Thus,$(x-2)$ is a factor.
By testing $f(-3) = 2(-27) - 3(9) - 17(-3) + 30 = -54 - 27 + 51 + 30 = 0$. Thus,$(x+3)$ is a factor.
Since $(x-2)$ and $(x+3)$ are factors,their product $(x-2)(x+3) = x^{2}+x-6$ is also a factor.
Dividing $2x^{3}-3x^{2}-17x+30$ by $(x^{2}+x-6)$:
$2x^{3}-3x^{2}-17x+30 = (x^{2}+x-6)(2x-5)$.
Further factoring $(x^{2}+x-6)$ into $(x-2)(x+3)$,we get:
$2x^{3}-3x^{2}-17x+30 = (x-2)(x+3)(2x-5)$.

(A) Let $f(x) = x^{3}-6x^{2}+11x-6$ be the given polynomial.
By the Factor Theorem,we test the factors of the constant term $-6$,which are $\pm 1, \pm 2, \pm 3, \pm 6$.
For $x = 1$,$f(1) = (1)^{3}-6(1)^{2}+11(1)-6 = 1-6+11-6 = 0$. Thus,$(x-1)$ is a factor.
For $x = 2$,$f(2) = (2)^{3}-6(2)^{2}+11(2)-6 = 8-24+22-6 = 0$. Thus,$(x-2)$ is a factor.
For $x = 3$,$f(3) = (3)^{3}-6(3)^{2}+11(3)-6 = 27-54+33-6 = 0$. Thus,$(x-3)$ is a factor.
Since the polynomial is of degree $3$,it can have at most $3$ linear factors.
Therefore,the factors of $x^{3}-6x^{2}+11x-6$ are $(x-1)(x-2)(x-3)$.

(B) Let $f(x) = x^{3} + x^{2} - 4x - 4$.
We can factorise by grouping terms:
$f(x) = (x^{3} + x^{2}) - (4x + 4)$
Factor out the common terms from each group:
$f(x) = x^{2}(x + 1) - 4(x + 1)$
Now,take $(x + 1)$ as a common factor:
$f(x) = (x + 1)(x^{2} - 4)$
Using the algebraic identity $a^{2} - b^{2} = (a - b)(a + b)$,we can factor $x^{2} - 4$ as $(x - 2)(x + 2)$:
$f(x) = (x + 1)(x - 2)(x + 2)$
Thus,the factors are $(x - 2)(x + 1)(x + 2)$.

(D) Let $f(x) = 3x^{3}-x^{2}-3x+1$ be the given polynomial.
We can factorize this by grouping terms:
$f(x) = (3x^{3}-3x) - (x^{2}-1)$
$f(x) = 3x(x^{2}-1) - 1(x^{2}-1)$
$f(x) = (3x-1)(x^{2}-1)$
Since $x^{2}-1$ is a difference of squares,we can write it as $(x-1)(x+1)$.
Therefore,$f(x) = (3x-1)(x-1)(x+1)$.

(N/A) To factorise the quadratic expression $9 x^{2}-12 x+3$,we use the splitting the middle term method.
First,we find two numbers whose product is $9 \times 3 = 27$ and whose sum is $-12$.
These two numbers are $-9$ and $-3$.
Now,rewrite the middle term $-12 x$ as $-9 x - 3 x$:
$9 x^{2}-9 x-3 x+3$
Group the terms:
$=9 x(x-1)-3(x-1)$
Factor out the common binomial $(x-1)$:
$=(9 x-3)(x-1)$
Finally,factor out the common constant $3$ from the first binomial:
$=3(3 x-1)(x-1)$

(A) The given expression is $1-64 a^{3}-12 a+48 a^{2}$.
We can rewrite this expression as $1^{3}-(4 a)^{3}-3(1)(4 a)(1-4 a)$.
This matches the algebraic identity $(x-y)^{3} = x^{3}-y^{3}-3xy(x-y)$,where $x=1$ and $y=4a$.
Therefore,the expression simplifies to $(1-4 a)^{3}$.
Thus,the factorised form is $(1-4 a)(1-4 a)(1-4 a)$.

(A) We have,
$1+64 x^{3} = (1)^{3} + (4 x)^{3}$
Using the algebraic identity $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$,where $a = 1$ and $b = 4x$:
$= (1 + 4x)((1)^{2} - (1)(4x) + (4x)^{2})$
$= (1 + 4x)(1 - 4x + 16x^{2})$
$= (1 + 4x)(16x^{2} - 4x + 1)$

(A) Area of the rectangle is given by the polynomial $4 a^{2}+4 a-3$.
To find the length and breadth,we factorize the quadratic expression by splitting the middle term.
We need two numbers whose sum is $+4$ and whose product is $4 \times (-3) = -12$.
These two numbers are $+6$ and $-2$,since $6 + (-2) = 4$ and $6 \times (-2) = -12$.
Splitting the middle term $4 a$ as $6 a - 2 a$,we get:
$4 a^{2} + 6 a - 2 a - 3$
Grouping the terms:
$= 2 a(2 a + 3) - 1(2 a + 3)$
Factoring out the common binomial $(2 a + 3)$:
$= (2 a - 1)(2 a + 3)$
Since the area of a rectangle is defined as $\text{length} \times \text{breadth}$,the possible expressions for the dimensions are:
Length $= (2 a - 1)$ and Breadth $= (2 a + 3)$ or Length $= (2 a + 3)$ and Breadth $= (2 a - 1)$.

(A) Let $f(x) = p x^{2} + 5 x + r$.
Since $(x-2)$ is a factor of $f(x)$,by the Factor Theorem,$f(2) = 0$.
$p(2)^{2} + 5(2) + r = 0$
$4p + 10 + r = 0$
$4p + r = -10$ --- $(1)$
Since $(x - \frac{1}{2})$ is a factor of $f(x)$,by the Factor Theorem,$f(\frac{1}{2}) = 0$.
$p(\frac{1}{2})^{2} + 5(\frac{1}{2}) + r = 0$
$\frac{1}{4}p + \frac{5}{2} + r = 0$
Multiply the entire equation by $4$:
$p + 10 + 4r = 0$
$p + 4r = -10$ --- $(2)$
Comparing equations $(1)$ and $(2)$,since both are equal to $-10$:
$4p + r = p + 4r$
$4p - p = 4r - r$
$3p = 3r$
$p = r$
Hence,it is proved.

(N/A) Let $p(x) = 2x^4 - 5x^3 + 2x^2 - x + 2$ and $g(x) = x^2 - 3x + 2$.
First,factorize the divisor $g(x)$:
$g(x) = x^2 - 3x + 2 = x^2 - x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)$.
For $p(x)$ to be divisible by $g(x)$,it must be divisible by both $(x - 1)$ and $(x - 2)$ by the Factor Theorem.
Check for $(x - 1)$:
$p(1) = 2(1)^4 - 5(1)^3 + 2(1)^2 - 1 + 2 = 2 - 5 + 2 - 1 + 2 = 0$.
Since $p(1) = 0$,$(x - 1)$ is a factor of $p(x)$.
Check for $(x - 2)$:
$p(2) = 2(2)^4 - 5(2)^3 + 2(2)^2 - 2 + 2 = 2(16) - 5(8) + 2(4) - 2 + 2 = 32 - 40 + 8 = 0$.
Since $p(2) = 0$,$(x - 2)$ is a factor of $p(x)$.
Since both $(x - 1)$ and $(x - 2)$ are factors of $p(x)$ and they are coprime,their product $(x - 1)(x - 2) = x^2 - 3x + 2$ must also be a factor of $p(x)$.
Hence,$2x^4 - 5x^3 + 2x^2 - x + 2$ is divisible by $x^2 - 3x + 2$.

(B) According to the Factor Theorem,$(x + 1)$ is a factor of $p(x)$ if and only if $p(-1) = 0$.
Given $p(x) = x^{4} + x^{3} + x^{2} + x + 1$.
Substituting $x = -1$ into the polynomial:
$p(-1) = (-1)^{4} + (-1)^{3} + (-1)^{2} + (-1) + 1$
Calculating each term:
$p(-1) = 1 - 1 + 1 - 1 + 1$
Simplifying the expression:
$p(-1) = 1$
Since $p(-1) \neq 0$,by the Factor Theorem,$(x + 1)$ is not a factor of the given polynomial.

(NONE) According to the Factor Theorem,$(x + 1)$ is a factor of a polynomial $p(x)$ if and only if $p(-1) = 0$.
Let $p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{4} + 3(-1)^{3} + 3(-1)^{2} + (-1) + 1$
$p(-1) = 1 + 3(-1) + 3(1) - 1 + 1$
$p(-1) = 1 - 3 + 3 - 1 + 1$
$p(-1) = 1$
Since $p(-1) = 1 \neq 0$,by the Factor Theorem,$(x + 1)$ is not a factor of the given polynomial.

(NONE) According to the Factor Theorem, $(x + 1)$ is a factor of a polynomial $p(x)$ if $p(-1) = 0$.
Let $p(x) = x^{3} - x^{2} - (2 + \sqrt{2})x + \sqrt{2}$.
Substitute $x = -1$ into the polynomial:
$p(-1) = (-1)^{3} - (-1)^{2} - (2 + \sqrt{2})(-1) + \sqrt{2}$
Calculate each term:
$p(-1) = -1 - 1 + (2 + \sqrt{2}) + \sqrt{2}$
$p(-1) = -2 + 2 + \sqrt{2} + \sqrt{2}$
$p(-1) = 2\sqrt{2}$
Since $p(-1) = 2\sqrt{2} \neq 0$, $(x + 1)$ is not a factor of the given polynomial.

(D) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if and only if $p(a) = 0$.
Here,$g(x) = x + 2$,so we set $x + 2 = 0$,which gives $x = -2$.
Now,we evaluate $p(-2)$:
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$
$p(-2) = -8 + 3(4) - 6 + 1$
$p(-2) = -8 + 12 - 6 + 1$
$p(-2) = 4 - 6 + 1 = -1$
Since $p(-2) \neq 0$,by the Factor Theorem,$g(x)$ is not a factor of $p(x)$.

(A) According to the Factor Theorem,$g(x) = x - a$ is a factor of $p(x)$ if $p(a) = 0$.
Here,$g(x) = x - 3$,so $a = 3$.
We calculate $p(3)$ by substituting $x = 3$ into $p(x) = x^3 - 4x^2 + x + 6$:
$p(3) = (3)^3 - 4(3)^2 + (3) + 6$
$p(3) = 27 - 4(9) + 3 + 6$
$p(3) = 27 - 36 + 3 + 6$
$p(3) = 36 - 36 = 0$
Since $p(3) = 0$,by the Factor Theorem,$g(x) = x - 3$ is a factor of $p(x) = x^3 - 4x^2 + x + 6$.

(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:
$1$. 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 $= 0$.

(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 $0$.

(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) 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.