Question

factorize
x3-3x2+3x+7

Answer 1

Remembering the standard formula for (a+b)3=a3+b3+3a2b+3ab2(a+b)3=a3+b3+3a2b+3ab2   will help in this question.

If a=x and b= 1, we get: (x+1)3=x3+1+3x2+3x(x+1)3=x3+1+3x2+3x 

We can notice that we already have all the terms except 1 in the given expression.

Let us add and subtract 1 in the expression:

x3+3x2+3x+1−1−7x3+3x2+3x+1−1−7 

= (x+1)3−8(x+1)3−8

= (x+1)3−23(x+1)3−23   (23=823=8)

Since a3−b3=(a−b)(a2+b2+ab)a3−b3=(a−b)(a2+b2+ab)

= (x+1−2)((x+1)2+22+(x+1)∗2)(x+1−2)((x+1)2+22+(x+1)∗2)

= (x−1)(x2+1+2x+4+2x+2)(x−1)(x2+1+2x+4+2x+2)

= (x−1)(x2+4x+7)(x−1)(x2+4x+7)

x2+4x+7x2+4x+7 can't be further factorized as it will result into non-real factors.

So, x3+3x2+3x+−7 =(x−1)(x2+4x+7)x3+3x2+3x+−7 =(x−1)(x2+4x+7)

PS: If you're looking for further factorization of quadratic term into complex  numbers domain, you can proceed it by using quadratic formula to find the roots and writing them as product of linear factors.

Answer 2

$\textbf{Answer -}$

Now, x³ - 3x² + 3x + 7

= x³ + x² - 4x² - 4x + 7x + 7

= x²(x + 1) + -4x (x + 1) + 7 (x + 1)

= (x + 1) (x² - 4x + 7),

which is the required factorization.

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