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Question
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 − 19x + 6
Solution
For x=2, the value of the given
expression `3x^3+2x^2-19x+6`
= `3(2)^3+2(2)^2-19(2)+6`
=`24+8-38+6`
= 0
⇒ x-2 is a factor of `3x^3+2x^2-19x+6`
Now let us do long division
Thus we have ,
`3x^3+2x^2-19x+6` =`(x-2) (3x^2+8x-3) `
=`(x-2)(3x^2+9x-x-3)`
=`(x-2)(3x(x+3)-(x+3))`
= (x-2) (3x-1) (x+3)
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