Question

Find the remainder, when f(y) = 4y3 – 12y2 + 11y – 5 is divided by g(y) = 2y – 1​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \:Find \: the \: remainder, \: when \: f(y) \: = {4y}^{3} - {12y}^{2} + 11y - 5$

$\sf \: is \: divided \: by \: g(y) = 2y - 1$

$\large\underline{\bold{Solution-}}$

Concept Used :-

Remainder Theorem :-

• This theorem states that if f(x) is a polynomial in x, then the remainder on dividing f(x) by x − a is f(a).

Let's solve the problem now!!

Now it is given that

$\sf \: g(y) = 2y - 1$

Put g(y) = 0, we get,

$\sf \: 2y - 1 = 0$

$\sf \: so \: y \: = \: \dfrac{1}{2}$

Now,

Given that,

$\sf \: f(y) \: = {4y}^{3} - {12y}^{2} + 11y - 5$

So, remainder when f(y) is divided by 2y - 1 is given by

$\sf \: f\bigg( \dfrac{1}{2} \bigg)$

$\sf \: = \: 4 {\bigg( \dfrac{1}{2} \bigg) }^{3} - 12 {\bigg( \dfrac{1}{2} \bigg) }^{2} + 11 {\bigg( \dfrac{1}{2} \bigg) } - 5$

$\sf \: = \: 4 \times \dfrac{1}{8} - 12 \times \dfrac{1}{4} + 11 \times \dfrac{1}{2} -$

$\sf \: = \: \dfrac{1}{2} - 3 + \dfrac{11}{2} - 5$

$\sf \: = \: 6 - 8$

$\sf \: = \: - \: 2$

Additional Information :-

Factor Theorem :-

• This theorem states that if f(x) is a polynomial in x then the remainder on dividing f(x) by x − a is 0, then x - a is factor of f(x).