Find the values of $p$ and $q$ so that $x^{4}+p x^{3}+2 x^{2}-3 x+q$ is divisible by $\left(x^{2}-1\right)$
Here, $f(x)=x^{4}+p x^{3}+2 x^{2}-3 x+q$
$g(x)=x^{2}-1$
first, we need to find the factors of $x^{2}-1$
$\Rightarrow x^{2}-1=0$
$\Rightarrow x^{2}=1$
$\Rightarrow x=\pm 1$
$\Rightarrow(x+1)$ and $(x-1)$
From factor theorem, if x = 1, -1 are the factors of f(x) then f(1) = 0 and f(-1) = 0
Let us take, x + 1
⟹ x + 1 = 0
⟹ x = -1
Substitute the value of x in f(x)
$f(-1)=(-1)^{4}+p(-1)^{3}+2(-1)^{2}-3(-1)+q$
= 1 - p + 2 + 3 + q
= -p + q + 6 .... 1
Let us take, x - 1
⟹ x - 1 = 0
⟹ x = 1
Substitute the value of x in f(x)
$f(1)=(1)^{4}+p(1)^{3}+2(1)^{2}-3(1)+q$
= 1 + p + 2 - 3 + q
= p + q ..... 2
Solve equations 1 and 2
- p + q = - 6
p + q = 0
2q = - 6
q = - 3
substitute q value in equation 2
p + q = 0
p - 3 = 0
p = 3
the values of are p = 3 and q = – 3
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.