Question

If the quotient on dividing $2x^4+x^3-14x^2-19x+6$ by $2x+1$ is $x^3+a{x}^2-bx-6$.
Find the values of a and b, also the remainder

Answer 1

Let $p\left(x\right)=2x^4+x^3-14x^2-19x+6$.

Given that the divisor is $2x+1$.

Now we get the value of x from

$2x+1=0$. Then $x=-\frac{1}{2}$

$\therefore$ The zero of the divisor is $x=-\frac{1}{2}$.
So, $2x^4+x^3-14x^2-19x+6=(x+\frac{1}{2})\{2x^3-14x-12\}+12$

$=(2x+1)\frac{1}{2}(2x^3-14x-12)+12$

Thus, the quotient is $\frac{1}{2}(2x^2-14x-12)=x^3-7x-6$ and the remainder is 12.

But, given quotient is $x^3+a{x}^2-bx-6$. Comparing this with the quotient obtained we get, a = 0 and b = 7.

Thus, a=0, b=7, and the remainder is 12.