Question

Using remainder theorum , factorise the expression 2x³ + x²- 13x + 6 completely​

Answer 1

Answer:

f(x) = 2x3 + x2 − 13x + 6 Factors of constant term 6 are ±1, ± 2, ± 3, ± 6. Putting x = 2, we have: f(2) = 2(2)3 + 22 − 13 (2) + 6 = 16 + 4 − 26 + 6 = 0 Hence (x − 2) is a factor of f(x). Read more on Sarthaks.com - https://www.sarthaks.com/150058/use-the-remainder-theorem-to-factorise-the-following-expression-2x-3-x-2-13x-6?show=150066#a150066

Answer 2

• Let f(x) = 2x³ + x² - 13x + 6
• Put x = 2
• f(2) = 2(2)³ + 2² - 13(2) + 6 = 0, satisfied.

Hence (x-2) is a factor of f(x) x-3

2x² + 5x -3

__________________

x - 2 ) 2x³ + x² - 13x + 6

2x³ - 4x²

- +

________________

5x² - 13x

5x² - 10x

- +

________________

- 3x + 6

- 3x + 6

+ -

____________

0

____________

→ f(x) = (x-2) (2x² + 5x - 3)

= (x - 2) (2x² + 6x - x - 3)

= (x -2) [2x ( x+3) - 1(x+3)]

= (x-2) (x +3) (2x-1).