Question

It is given that 2x³-7x²+x+10 = (x - 1) Q(j) + R. Find the remainder R and polynomial Q(j).​

Answer 1

`\large \text{$ Q(j) = 2x^2 - 5x - 4 $}`

`\large \text{R = 6}`

Explanation:

Use long division to find the polynomial and remainder:

`\large \begin{array}{r}2x^2-5x-4\phantom{)}\\x-1{\overline{\smash{\big)}\,2x^3-7x^2+x+10\phantom{)}}}\\\underline{-~\phantom{(}(2x^3-2x^2)\phantom{-b)))))))}}\\0-5x^2+x+10\phantom{)}\\\underline{-~\phantom{()}(-5x^2+5x)\phantom{-b))}}\\0-4x+10\phantom{)}\\\underline{-~\phantom{()}(-4x+\phantom{)}4)}\\6\phantom{)}\end{array}`

Therefore:

`\large \text{$2x^3 - 7x^2 + x + 10 = (x - 1)(2x^2 - 5x - 4) + 6$}`

So the polynomial and remainder are:

• 
  `\large \text{$ Q(j) = 2x^2 - 5x - 4 $}`
• 
  `\large \text{R = 6}`

Answer 2

• 2x²- 5x - 4, remainder 6

Explanation:

Divide the given by x - 1:

x - 1 | 2x³ - 7x² + x + 10 = 2x²- 5x - 4 rem 6

| 2x³ - 2x²

-5x² + x

-5x² + 5x

- 4x + 10

- 4x + 4

6

2x³-7x²+x+10 = (x - 1)(2x²- 5x - 4) + 6