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Question
Using the Remainder Theorem, factorise the following completely:
3x3 + 2x2 – 19x + 6
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Solution
Let P(x) = 3x3 + 2x2 – 19x + 6
By hit and trial method,
P(1) = 3(1)3 + 2(1)2 – 19(1) + 6
= 3 + 2 – 19 + 6
= –8 ≠ 0
P(–1) = 3(–1)3 + 2(–1)2 – 19(–1) + 6
= –3 + 2 + 19 + 6
= 24 ≠ 0
P(2) = 3(2)3 + 2(2)2 – 19(2) + 6
= 24 + 8 – 38 + 6
= 0
Thus, (x – 2) is a factor of P(x).
Now,
3x2 + 8x – 3
`x - 2")"overline(3x^3 + 2x^2 - 19x + 6)`
3x3 – 6x2
– +
8x2 – 19x + 6
8x2 – 16x
– +
– 3x + 6
– 3x + 6
+ –
0
∴ 3x3 + 2x2 – 19x + 6 = (x – 2)(3x2 + 8x – 3)
= (x – 2)(3x2 + 9x – x – 3)
= (x – 2)(3x(x + 3) – 1(x + 3))
= (x – 2)(x + 3)(3x – 1)
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