# find the real, linear, quadratic and irreducible factors of x^5 - 32

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x^5-32. We have to factorise this.

Solution:

Let f(x) = x^5- 32.

Put x= 2 in f(x) = x^5- 32.

f(2) = 2^5 - 32 = 32-32 = 0.

So by remainder theorem, (x-2) is a factor of x^5-32.

We know that (x^n-a^n)/(x-a) = x^(n-1) - ax^(n-2)++a^2*x^(n-3)+a^3*x^(n-4)+....+a^(n-2)*x +a^n is an identity.Therefore (x^n-a^n) = (x-a) {x^(n-1) - ax^(n-2)++a^2*x^(n-3)+a^3*x^(n-4)+....+a^(n-2)*x +a^n}

We use the above to factor f(x) = x^5 - 32 = x^5 - 2^5..

Therefore x^45 - 2^5 = (x-2){ x^4+2x^3+2^2*x^2+2^3x+2^4}.

x^5- 32 = (x-2)(x^4+2x^3+4x^2+8x+16).

So x-2 is the linear factor.

**So x^5 - 32 = (x-2)(x^4+2x^3+4x^2+8x+16). **