Factorising polynomials with complex roots

Factor theorem

Using the factor theorem, we know that if a is a factor of f(x), one of the factors of f(x) is x-a. This also works for polynomials with complex roots.

If a+bi is a root, that means that (x-[a+bi]), otherwise written as (x-a-bi), is a factor.

Factorising polynomials with complex roots

To factorise a polynomial which has a complex root (i.e. the discriminant is negative), we can first find the complex roots by solving the polynomial equal to zero, and then use the factor theorem to write the factors.

That’s because if r is a root of f(x), then (x-r) is a factor of f(x) - as we learned just above with the factor theorem.

Factorise x^2-4x+40 by first solving

• Let x^2-4x+40=0
• (x-2)^2-4+40=0
• (x-2)^2=-36
• x-2=\sqrt{-36}
• x=2\pm6i
• Roots: x=2+6i or x=2-6i
• Factors (by the factor theorem): [x-(2+6i)] and [x-(2-6i)]
• Amnswer: [x-(2+6i)][x-(2-6i)]

Factorise x^2+2x+5 by first solving

• Let x^2+2x+5=0
• (x+1)^2-1+5=0
• (x+1)^2= -4
• x+1=\sqrt{-4}
• x=-1\pm2i
• Roots: x=-1+2i or x=-1-2i
• Factors (by the factor theorem): [x-(-1+2i)] and [x-(-1-2i)]
• Answer: [x-(-1+2i)][x-(-1-2i)]

Factorise 2x^2+4x+10 by first solving

• Let 2x^2+4x+10=0
• 2(x^2+2x+5)=0
• (x+1)^2-1+5=0
• (x+1)^2= -4
• x+1=\sqrt{-4}
• x=-1\pm2i
• Roots: x=-1+2i or x=-1-2i
• Factors (by the factor theorem): [x-(-1+2i)] and [x-(-1-2i)]
• Answer: 2[x-(-1+2i)][x-(-1-2i)]

Factorise x^2-6x+13 by first solving

• Let x^2-6x+13=0
• (x-3)^2-9+13=0
• (x-3)^2= -4
• x-3=\sqrt{-4}
• x=3\pm2i
• Roots: x=3+2i or x=3-2i
• Factors (by the factor theorem): [x-(3+2i)] and [x-(3-2i)]
• Answer: [x-(3+2i)][x-(3-2i)]

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