the polynomial p(x) = $\displaystyle x^4+ax^3-10x^2+bx-12$ is divisible by d(x) = $\displaystyle (x+4)(x-3)$ Calculate the quotient of the division of P(x) of D(x).
the polynomial p(x) = $\displaystyle x^4+ax^3-10x^2+bx-12$ is divisible by d(x) = $\displaystyle (x+4)(x-3)$ Calculate the quotient of the division of P(x) of D(x).
Answer:
$\displaystyle x^2+x+1$
Hi
Dividing $\displaystyle x^4+ax^3-10x^2+bx-12$ by d(x) = $\displaystyle (x+4)(x-3)$ gives
$\displaystyle x^2+(a-1)x+3-a$ and a rest $\displaystyle (b+13a-15)x+12 (2-a)$
The rest is equal to 0 therefore
a = 2
b = -11
Quotient is $\displaystyle x^2+x+1$