Question

Use remainder factor theorum to factorise
$\sf2 {x}^{3} + {x}^{2} - 13x + 6$
​

Answer 1

Answer:

Let f(x)=2x

3

+x

2

–13x+6

For x=2, the value of f(x) will be

f(2)=2(2)

3

+(2)

2

–13(2)+6=16+4–26+6=0

As f(2)=0, so (x–2) is a factor of f(x).

Now, performing long division we have

Thus, f(x)=(x−2)(2x

2

+5x–3)

=(x–2)[2x

2

+6x–x–3]

=(x–2)[2x(x+3)−1(x+3)]

=(x–2)[2x(x+3)−1(x+3)]

=(x–2)(2x–1)(x+3)