factorise 2x³+3x²-5x-6 using remainder theorum
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Answered by
1
Let f(x)=2x
3
+3x
2
−9x−10, then f(−1)=−2+3+9−10=0
So, by remainder theorem, when f(x) is divided by x+1, the remainder is 0. So, (x+1) is a factor of f(x)
f(2)=16+12−18−10=0
So, by similar reasoning, (x−2) is also a factor of f(x)
Therefore, f(x)=(x+1)(x−2)p(x)=(x
2
−x−2)p(x)
By synthetic division, we can find p(x)
⇒f(x)=2x
3
+3x
2
−9x−10=(x
2
−x−2)(2x+5)
Therefore p(x)=2x+5 and the factorization of f(x) is
f(x)=(x+1)(x−2)(2x+5)
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