Question

Find the value of p and q in the polynomial $\text{f(x)}={\text{x}}^3-{\text{px}}^2+14\text{x}–\text{q}$, if it is exactly divisible by $\text{(x-1)}$ and $\text{(x-2)}$.​

Answer 1

Step 1: Form the equations

Given: $\text{f(x)}={\text{x}}^3-{\text{px}}^2+14\text{x}–\text{q}$

A polynomial $\text{f(x)}$ is exactly divisible by $\text{(x-a)}$ if and only if $\text{f(a)}=0$.

$\text{f(x)}$ is exactly divisible by $\text{(x-1)}$

$\therefore \text{f(1)}=0$
$(1{)}^3-\text{p}(1{)}^2+14\left(1\right)-\text{q}=0$
$1-\text{p}+14-\text{q}=0$
$p=15-\text{q}$ (i)

Also, f(x) is exactly divisible by $\text{f(x-2)}$.
$\therefore \text{f(2)}=0$
$(2{)}^3-\text{p}(2{)}^2+14\left(2\right)-\text{q}=0$
$8-4\text{p}+28-\text{q}=0$
$4\text{p}+\text{q}=36$ (ii)

Step 2: Find the value of p and q

Put the value of p in equation (ii)
$4(15-\text{q})+\text{q}=36$
$60-4\text{q}+\text{q}=36$
$3\text{q}=24$
$\text{q}=8$

Put the value of q in equation ()
$\text{p}=15-8$
$\text{p}=7$

Hence, $\text{p}=7$ and $\text{q}=8$.