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Question
Using the Remainder Theorem, factorise the following completely:
2x3 + x2 – 13x + 6
Solution
Let f(x) = 2x3 + x2 – 13x + 6
For x = 2
Factors of constant term 6 are ±1, ±2, ±3, ±6.
Putting x = 2, we have:
f(x) = f(2)
= 2(2)3 + (2)2 – 13(2) + 6
= 16 + 4 – 26 + 6
= 0
Hence, (x – 2) is a factor of f(x).
2x2 + 5x – 3
`x - 2")"overline(2x^3 + x^2 - 13x + 6)`
2x3 – 4x2
– +
5x2 – 13x
5x2 – 10x
– +
– 3x + 6
– 3x + 6
+ –
0
∴ 2x3 + x2 – 13x + 6 = (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)
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