Factoring

by constant and/or variable(s)

Let’s start with some examples

a) \(6x^2-x=x(6x-1)\)

b) \(-8x^4+6=2(-4x^4+3)=-2(4x^4-3)\) Here both are considered valid answers.

c) \(5xy^2-xy+9y=y(5xy-x+9)\)

d) \(\frac{9}{2}x^2+\frac{1}{2}x=x(\frac{9}{2}x+\frac{1}{2})=\frac{1}{2}x(9x+1)\)

e) \(-10a^2b^2+6a^2b-ab=ab(-10ab+6a-1)\)

f) \(20x^3+16x^2+4=4(5x^3+4x^2)+4=4(5x^3+4x^2+1)=4(x+1)(5x^2-x+1)\) Only products are valid answers.

Factoring by division

for quadratic and higher degree of polynomials

Here is a link for the technique used in Finnish schools: Polynomin jakaminen polynomilla jakokulmassa | Opetus.tv

Examples

a) \(P(x)=5x^2-10x-15\)

Let’s try to find a value for the variable x so that \(P(x)=0\).

It turns out that if \(x=-1\), then \(P(x)=0\). This means that one factor of the polynomial \(P(x)\) is \((x+1)\).

Then there are different techniques or algorithms to find out what the other factor will be. I (the author) use a method called a “division angle” since it is an algorithm and it doesn’t need “trial and error”.

We found that the second factor is \((5x-15)\), so the polynomial can be factorized as

\(P(x)=(x+1)(5x-15)\)

But this can be factorized even further! The final answer is \(P(x)=5(x+1)(x-3)\) so always remember to factorize as far as possible. With polynomials of higher degree, you have to continue to find out other factors as well.

b) \(P(x)=-x^2+4x-3\)

One solution for \(P(x)=0\) would be \(x=1\), since \(P(1)=-1^2+4\cdot 1-3=0\). Therefore one factor of \(P(x)\) is \(x-1\). Let’s find the oher factor.

So \(P(x)=-x^2+4x-3=(x-1)(-x+3)\).

Polynomials with higher degree

Same algorithm works here: if one value for the variable \(x\) is \(k\) so that \(P(k)=0\), then the polynomial with a degree of \(n\) will be factored to \((x-m)\) and to a polynomial with a degree of \(n-1\). Then continue to factor out the polynomial with a degree of \(n-1\) as well.

Examples

a) \(P(x)=x^3+2x^2+3x+6\)

b) \(Q(x)=2x^3+5x^2-36x+36\)

c) \(R(x)=4x^4+14x^3-62x^2+72\)

Special forms of polynomials

Here are some useful special forms of polynomials:

\(x^2+2xy+y^2=(x+y)^2\) which means that usually \(x^2+y^2\ne(x+y)^2\)!

\(x^2-2xy+y^2=(x-y)^2\)

\(x^2-y^2=(x+y)(x-y)\)

\(x^3+y^3=(x+y)(x^2-xy+b^2)\)

\(x^3-y^3=(x-y)(x^2+xy+y^2)\)

Examples

a) \(A(x)=6x^2+12x+6\)

b) \(B(x)=x^2+10xy+25y^2\)

c) \(C(y)=4y^2-16y+16\)

d) \(D(x)=16x^2-9\)

e) \(E(x)=25-x^2\)

f) \(F(x)=8x^3+27\)

g) \(R(x)=4x^4+14x^3-62x^2+72\)

Division of polynomials

Division is done by factoring the numerator and/or the denominator in order to find something to simplify from both sides of the quotient. The algorithm of division angel can in some cases be useful. If the division isn’t even, what will be left is a remainder.

For eaxmple

\(\frac{(3x^3-7x^2+5x-6)}{(x-2)}=\frac{(x-2)(3x^2-x+3)}{(x-2)}=3x^2-x+3\)

Finally factoring by gouping

There are no neat algorithms or tricks to these examples. Just some examples of tougher cases. 🤨

\(\begin{align}\text{a)} \ 5x^2+9x-2&=5x^2-x+10x-2 \\ &=x(5x-1)+2(5x-1) \\ &=(5x-1)(x+2)\end{align}\)

\(\begin{align}\text{b)} \ 2x^4-3x^3+2x-3&=2x^4+2x-3x^3-3 \\ &=2x(x^3+1)-3(x^3+1) \\ &=(x^3+1)(2x-3)\end{align}\)

\(\begin{align}\text{c)} \ x^5-2x^3-4x^2+8&=x^3(x^2-2)-4(x^2-2) \\ &=(x^2-2)(x^3-4)\end{align}\)