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.

The Square of a Binomial

\(\begin{align} I) \ (x+y)^2&=x^2+2xy+y^2, \ \text{which means that usually that} \ x^2+y^2\ne(x+y)^2 \\ II) \ (x-y)^2&=x^2-2xy+y^2 \\ \\ III) \ x^2-y^2&=(x+y)(x-y) \\ IV) \ x^3+y^3&=(x+y)(x^2-xy+b^2) \\ V) \ x^3-y^3&=(x-y)(x^2+xy+y^2 \end{align}\)

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

Factoring by gouping

Factoring of quadratic polynomials of the form of \(P(x)=ax^2+c\) can be factored, if possible, by using the special form of \(x^2-y^2=(x+y)(x-y)\).

For complete quadratic polynomials of the form of \(P(x)=ax^2+bx+c\) a technic called grouping is an effective way. For example, let \(P(x)=5x^2+9x-2\). Now we have to solve a pair of equations that satisfies these two conditions:

When \(ab=5\cdot(-2)=-10\) and \(b=9\), then

\(\begin{align} kl&=-10 \\ k+l&=9\end{align}\)

To work this out, we have to find such integers \(k\) and \(l\) so that their product equals -10 and their sum equals 9. We can do this by writing down all the possible combinations of \(k \cdot l\) that produces -10.

\(\begin{align} 1\cdot(-10)&=-10 \\ -1\cdot10&=-10 \\ 2\cdot(-5)&=-10 \\ -2\cdot5&=-10\end{align}\)

From the combinations we can see that \(-1+10\) is equal to \(9\), so that is our solution. Now we can write \(\textbf{9x = -x + 10x}\) and substitute that in the polynomial.

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

Another example

Let \(Q(x)=5x^2+11x-12\). This time \(k \cdot l=5\cdot(-12)=-60\)

\(\begin{align} -1\cdot60&=-60 \\ -2\cdot30&=-60 \\ -3\cdot20&=-60 \\ -4\cdot15&=-60 \end{align}\)

From the last combination we can now pick \(k+l=-4+15=11\). Therefore

\(\begin{align}5x^2+11x-12&=5x^2-4x+15x-12 \\ &=5x^2+15x-4x-12 \ (\text{This time the order of} \ k \ \text{and} \ l \ \text{had to be changed}) \\ &=5x(x+3)-4(x+3) \\ &=(x+3)(5x-4) \end{align}\)

Here is little video of the solution.

And finally, couple of harder examples to look into.

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

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