if p(x)=x^3-4x^2+x+6 then show that p(3)=0 and hence factorize p(x)
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Answer:
p(x) = x^3 - 4x^2 + x + 6
=> p (3) = ( 3)^3 - 4(3)^2 + (3) + 6
=> p(3) = 27 - 36 + 3 + 6
=> p(3) = 27 - 36 + 9
=> p(3) = 27 + 9 - 36
=> p(3) = 36 - 36
=> p(3) = 0
Now,
x^3 - 4x^2 + x + 6
= x^3 - 2x^2 - 2x^2 + 4x - 3x + 6
= x^2 ( x - 2) - 2x ( x - 2) - 3(x - 2)
= ( x - 2) (x^2 - 2x - 3)
= (x-2) [ x^2 - 3x + x - 3]
= (x - 2) [ x (x - 3) + 1 (x - 3)]
= (x-2)(x-3) (x+1)
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