Factors of Polynomials (Cambridge International IGCSE Further Maths): Exam Questions

$p(x)=15x^{3}+22x^{2}-15x+2$

Find the remainder when $p(x)$ is divided by $x+1$.

Extract:

p(x) = 15x³ + 22x² - 15x + 2

(i) Show that $x+2$ is a factor of $p(x)$ .

(ii) Write $p(x)$ as a product of linear factors.

The polynomial $p(x)=6x^{3}+ax^{2}+bx+2$, where $a$ and $b$ are integers, has a factor of $x--2$ . Given that $p(1)=--2p(0)$, find the values of $a$ and $b$.

Extract:

The polynomial p(x) = 6x³ + ax² + bx + 2, where a and b are integers, has a factor of x - 2.

Using your values of $a$ and $b$,

(i) find the remainder when $p(x)$ is divided by$2x--1$

(ii) factorise $p(x)$.

$p(x)=ax^{3}+3x^{2}+bx-12$ has a factor of $2x+1$. When $p(x)$ is divided by $x-3$ the remainder is 105.

Find the value of $a$ and of $b$.

Extract:

p(x) = ax³ + 3x² + bx - 12 has a factor of 2x + 1.

a = 6, b = -24

Using your values of $a$ and $b$, write $p(x)$ as a product of $2x+1$ and a quadratic factor.

Extract:

Using your values of a and b, write p(x) as a product of 2x + 1 and a quadratic factor.

a = 6, b = -24 so p(x) = 6x³ + 3x² - 24x - 12

Rearrange p(x) so that the first 2 terms have a common factor of 2x and the second 2 terms have a common factor of 1

p(x) = 6x³ - 24x + 3x² - 12

Take out the common factor of the first 2 terms and the second 2 terms

p(x) = 2x(3x² - 12) + 1(3x² - 12)

Factorise

p(x) = (2x + 1)(3x² - 12)

Hence solve$p(x)$= 0.

$p(x)=6x^{3}+ax^{2}+12x+b$, where $a$ and $b$ are integers.

$p(x)$ has a remainder of 11 when divided by $x-3$ and a remainder of $-21$ when divided by $x+1$. Given that $p(x)=(x-2)Q(x)$, find $Q(x)$, a quadratic factor with numerical coefficients.

Extract:

p(x) = 6x³ + ax² + 12x + b, where a and b are integers.

p(x) has a remainder of 11 when divided by x - 3 and a remainder of -21 when divided by x + 1.

Given that p(x) = (x - 2)Q(x), find Q(x), a quadratic factor with numerical coefficients.

Using remainder theorem and solving simultaneous equations: a = -23 and b = 20.

Therefore p(x) = 6x³ - 23x² + 12x + 20

Using algebraic long division:

p(x) = (x - 2)(6x² - 11x - 10)

Hence solve $p(x)=0$.

The three roots of $\mathrm{p} ( x ) = 0$ , where $\mathrm{p} ( x ) = 2 x^{3} + a x^{2} + b x + c$ are $x=\frac{1}{2},x=n$ and $x=-n$, where $a,b,c$ and $n$ are integers. The $y$-intercept of the graph of $y=p(x)$ is 4. Find $p(x)$, simplifying your coefficients.

The polynomial $p(x)=ax^{3}-9x^{2}+bx-6$, where $a$ and $b$ are constants, has a factor of $x-2$. The polynomial has a remainder of 66 when divided by $x-3$. Find the value of $a$ and of $b$.

Extract:

The polynomial p(x) = ax³ - 9x² + bx - 6, where a and b are constants, has a factor of x - 2. The polynomial has a remainder of 66 when divided by x - 3.

a = 6 and b = -3

Using your values of $a$ and $b$, show that $p(x)=(x-2)q(x)$, where $q(x)$ is a quadratic factor to be found.

Extract:

Using your values of a and b, show that p(x) = (x - 2)q(x), where q(x) is a quadratic factor to be found.

To work out the quadratic factor q(x), we can use algebraic long division to divide p(x) by x - 2

q(x) = 6x² + 3x + 3

Hence show that the equation $p(x)=0$ has only one real solution.

$p(x)=2x^{3}-3x^{2}-23x+12$

Find the value of $\mathrm{p} ( 1 \frac{1}{2} )$.

Extract:

p(x) = 2x³ - 3x² - 23x + 12

Find the value of p(1/2).

Write $p(x)_{}$ as the product of three linear factors and hence solve $p(x)=0$.

The polynomial $p(x)=ax^{3}+bx^{2}-19x+4$, where $a$ and $b$ are constants, has a factor $x+4$ and is such that $2p(1)=5p(0)$.

Show that $p(x)=(x+4)(Ax^{2}+Bx+C)$, where $A$, $B$ and $C$ are integers to be found.

Extract:

The polynomial p(x) = ax³ + bx² - 19x + 4, where a and b are constants, has a factor x + 4 and is such that 2p(1) = 5p(0).

From part a, we found that a = 6 and b = 19, giving: p(x) = 6x³ + 19x² - 19x + 4

And we showed that: p(x) = (x + 4)(Ax² + Bx + C)

where A = 6, B = -5, and C = 1.

Therefore: p(x) = (x + 4)(6x² - 5x + 1)

Hence factorise $p(x)$.

Extract:

The polynomial p(x) = ax³ + bx² - 19x + 4, where a and b are constants, has a factor x + 4 and is such that 2p(1) = 5p(0).

From the mark scheme: a = 6 b = 19

Substituting these values into the polynomial, p(x) = 6x³ + 19x² - 19x + 4

Find the remainder when $p'(x)$ is divided by$x$.