Question
Using the Remainder Theorem, factorise completely the following polynomial:
$3x^2 + 2x^2 – 19x + 6$
$3x^2 + 2x^2 – 19x + 6$
✓
Answer
Let $f(x) = 3x^2 + 2x^2 – 19x + 6$
Using hit and trial method,
$f(1) = 3 + 2 – 19 + 6 \neq 0$
$f(−1) =–3 + 2 + 19 + 6 \neq 0$
$f(2) = 24 + 8 – 38 + 6 = 0$
Hence, (x – 2) is a factor of f(x)
To factorise $3x^2+ 8x − 3$
$= 3x^2+ 9x − x − 3$
$= 3x(x + 3) −1(x + 3)$
$= (3x − 1)(x + 3)$
Hence $3x^3+ 2x^3−19x + 6 = (x − 2)(3x − 1)(x + 3)$
Using hit and trial method,
$f(1) = 3 + 2 – 19 + 6 \neq 0$
$f(−1) =–3 + 2 + 19 + 6 \neq 0$
$f(2) = 24 + 8 – 38 + 6 = 0$
Hence, (x – 2) is a factor of f(x)
To factorise $3x^2+ 8x − 3$
$= 3x^2+ 9x − x − 3$
$= 3x(x + 3) −1(x + 3)$
$= (3x − 1)(x + 3)$
Hence $3x^3+ 2x^3−19x + 6 = (x − 2)(3x − 1)(x + 3)$
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