Factors of Polynomials (Cambridge (CIE) IGCSE Additional Maths): Exam Questions

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

Using your values of $a$ and $b$,

(i) find the remainder when $\mathrm{p}\left(x\right)$ is divided by$2x–1$

(ii) factorise $\mathrm{p}\left(x\right)$.

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

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

Using your values of $a$ and $b$, write $\mathrm{p}\left(x\right)$ as a product of $2x+1$ and a quadratic factor.

Hence solve$\mathrm{p}\left(x\right)$= 0.

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

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

Hence solve $\mathrm{p}\left(x\right)=0$.

The three roots of $\mathrm{p}\left(x\right)=0$ , where $\mathrm{p}\left(x\right)=2x^3+a{x}^2+bx+c$ 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 $y=\mathrm{p}\left(x\right)$ is 4. Find $\mathrm{p}\left(x\right)$, simplifying your coefficients.

The polynomial $\mathrm{p}\left(x\right)=a{x}^3-9x^2+bx-6$, 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 $b$.

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

Hence show that the equation $\mathrm{p}\left(x\right)=0$ has only one real solution.

$\mathrm{p}\left(x\right)=2x^3-3x^2-23x+12$

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

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

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

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

Hence factorise $\mathrm{p}\left(x\right)$.

Find the remainder when $\mathrm{p}\text{'}\left(x\right)$ is divided by$x$.