Question

factorise= (x^3+3x^2+3x-7)​

Answer 1

Answer:

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.

Step-by-step explanation:

Hope this helped ≧ω≦