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Question
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 – 23x – 30
Solution
`f(x)=3x^3+2x^2-23x-30`
for x=-2,
`f(x)=f(-2)=3(-2)^3+2(2)^2-23(-2)-30`
=`-24+8+46-30=-54+54=0`
Hence, (x+2)is a factor of f(x).
∴ `3x^3+2x^2-23x-30=(x+2)(3x^2+4x-15)`
=`(x+2)(3x^2+5x-9x-15)`
= `(x+2) [x(3x+5)-3(3x+5)]`
= `(x-2) (3x+5)(x-3)`
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