Factorising from roots (includes complex roots)

Suppose the polynomial $P\left(x\right)$P(x) is a polynomial with real coefficients. If the corresponding polynomial equation $P\left(x\right)=0$P(x)=0 of degree $n$n has $n$n real roots, then $P\left(x\right)$P(x) can be factorised over the reals into $n$n linear factors.

If $P\left(x\right)=0$P(x)=0 has less than $n$n real roots, then there must also exist conjugate pairs of complex roots of the form $a\pm bi$a±bi.  This is a consequence of the conjugate root theorem which tells us that if $z_1=a+bi$z1​=a+bi is a root of a  polynomial equation  $P\left(x\right)=0$P(x)=0 with real coefficients, then so also is $z_2=a-bi$z2​=a−bi.

For example the degree $4$4 polynomial equation given by $x^4-2x^3+2x^2+38x-39=0$x4−2x3+2x2+38x−39=0 has two real roots $x=1$x=1 and $x=-3$x=−3 , and two complex conjugate roots $2\pm3i$2±3i. 

This means therefore that the polynomial $P\left(x\right)=x^4-2x^3+2x^2+38x-39$P(x)=x4−2x3+2x2+38x−39  can be factorised as $P\left(x\right)=\left(x-1\right)\left(x+3\right)\left(x-2-3i\right)\left(x-2+3i\right)$P(x)=(x−1)(x+3)(x−2−3i)(x−2+3i). 

Example 1

Suppose we needed to factorise the polynomial $P\left(x\right)=x^4+x^3-7x^2-x+6$P(x)=x4+x3−7x2−x+6 given that we know that $\left(x+3\right)$(x+3) is a factor.

Then we can use synthetic division as follows:

$P\left(x\right)$P(x)	$1$1	$1$1	$-7$−7	$-1$−1	$6$6
$-3$−3		$-3$−3	$6$6	$3$3	$-6$−6
$Q\left(x\right)$Q(x)	$1$1	$-2$−2	$-1$−1	$2$2	$0$0

Hence $P\left(x\right)=\left(x+3\right)\left(x^3-2x^2-x+2\right)$P(x)=(x+3)(x3−2x2−x+2). 

We may notice that the quotient polynomial $Q\left(x\right)=x^3-2x^2-x+2$Q(x)=x3−2x2−x+2  can be factorised in two stages. Firstly, $Q\left(x\right)=x^2\left(x-2\right)-\left(x-2\right)$Q(x)=x2(x−2)−(x−2) and so taking out the common factor we see that $Q\left(x\right)=\left(x-2\right)\left(x^2-1\right)=\left(x-2\right)\left(x-1\right)\left(x+1\right)$Q(x)=(x−2)(x2−1)=(x−2)(x−1)(x+1).

This means that $P\left(x\right)=\left(x+3\right)\left(x-2\right)\left(x-1\right)\left(x+1\right)$P(x)=(x+3)(x−2)(x−1)(x+1). 

The zeros of $P\left(x\right)=0$P(x)=0 are then given as $-3,2,-1,1$−3,2,−1,1. 

Example 2

Suppose we know that the polynomial $P\left(x\right)=x^3-3x^2+49x-147$P(x)=x3−3x2+49x−147 has the factor $\left(x-7i\right)$(x−7i).

Then by the conjugate root theorem, $P\left(x\right)$P(x) must also contain the factor $\left(x+7i\right)$(x+7i). This means that $P\left(x\right)=\left(x-a\right)\left(x-7i\right)\left(x+7i\right)=\left(x-a\right)\left(x^2+49\right)$P(x)=(x−a)(x−7i)(x+7i)=(x−a)(x2+49), for some real number a.

If we were to multiply out these factors, the constant term $-49a$−49a would have to equal $-147$−147. Thus $a=3$a=3 and so $P\left(x\right)=\left(x-3\right)\left(x-7i\right)\left(x+7i\right)$P(x)=(x−3)(x−7i)(x+7i).

The zeros of $P\left(x\right)=0$P(x)=0 are then given as $3,7+i,7-i$3,7+i,7−i. 

Example 3

The polynomial $P\left(x\right)=x^4-2x^3+6x^2-8x+8$P(x)=x4−2x3+6x2−8x+8 has the factor $\left(x-2i\right)$(x−2i).

Again from the conjugate root theorem it must also have the factor $\left(x+2i\right)$(x+2i), and so we can express the polynomial as $P\left(x\right)=\left(x^2+4\right)Q\left(x\right)$P(x)=(x2+4)Q(x) where $Q\left(x\right)$Q(x) is a polynomial of degree $2$2. 

By division we can show that $Q\left(x\right)=x^2-2x+2$Q(x)=x2−2x+2. 

Noting that $\left(x-1\right)^2=x^2-2x+1$(x−1)2=x2−2x+1, we see that $Q\left(x\right)=\left(x-1\right)^2+1$Q(x)=(x−1)2+1 and this can only be factorised over the complex field.

Thus $Q\left(x\right)=\left(x-1\right)^2+1=\left(x-1\right)^2-i^2=\left(x-1-i\right)\left(x-1+i\right)$Q(x)=(x−1)2+1=(x−1)2−i2=(x−1−i)(x−1+i).

Hence $P\left(x\right)=\left(x+2i\right)\left(x-2i\right)\left(x-1-i\right)\left(x-1+i\right)$P(x)=(x+2i)(x−2i)(x−1−i)(x−1+i).

The zeros of $P\left(x\right)=0$P(x)=0 are then given as $2i,-2i,1+i,1-i$2i,−2i,1+i,1−i. 

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3