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Question
Use factor theorem to factorise: 3x3 + 2x2 − 19x + 6.
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Solution
Let p(x) 3x3 + 2x2 − 19x + 6
Factors of constant term + 6 are ±1, ±2, ±3, ±6
Put x = 1;
p(1) = 3(1)3 + 2(1)2 − 19(1) + 6
= 3 + 2 − 19 + 6
= −8
So (x – 1) is not a factor.
Put x = 2
p(2) = 3(2)3 + 2(2)2 − 19(2) + 6
= 3(8) + 2(4) − 19(2) + 6
= 24 + 8 − 38 + 6
= 0
x − 2 is factor of p(x).
3x2 + 8x − 3
`x − 2")"overline(3x^3 + 2x^2 - 19x + 6)`
3x3 − 6x2
− +
8x2 − 19x
8x2 − 16x
− +
− 3x + 6
− 3x + 6
+ −
x
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)
