# Factoring Polynomials With Large Coefficients: Factoring by Extraction

Factoring polynomials, in general, is quite difficult, but some special ones can be factored using certain tricks.

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## Factorizing Quadratics with Large Numbers

Today, I will discuss how to factor polynomials with large coefficients such as \(3x^2+10x-100\) with ease. I know that this will be a long note but I feel that it is worth reading everything including the generalized form at the bottom except for the proof (unless you want to).

While sitting in my math class today, I discovered a trick to factoring second-degree polynomials with large or irrational second and third coefficients. For example, try factoring \(3x^2+10x-1000\). It's relatively simple to factor it to \((3x-50)(x+20)\) but that would take a little while or at least longer than the way that I'm about to discuss.

We begin with the expression \(3x^2+10x-1000\). Then we divide the second coefficient by 10 and the third by 100, and we are left with the expression \(3x^2+x-10,\) which we can easily factor to \((3x-5)(x+2)\). Finally, we multiply the second term in each factor by 10 and we have \((3x-50)(x+20)\). Looks familiar, doesn't it?

*Basically, what I have done is that I divided the second coefficient by any one of its factors (in this case 10) and then divided the third coefficient by the square of that factor while leaving the first untouched.*

This method applies to irrational and imaginary coefficients as well.

Factorize \(4x^2+8\sqrt2x+8\).

Factor out \(2\sqrt2\) from the second coefficient and 8 from the third, and then we are left with \(4x^2+4x+1=(2x+1)^2.\) By multiplying back \(2\sqrt2\) to the second term in each factor which in this case is \((2x+1),\) we have

\[\left(2x+2\sqrt2\right)^2. \ _\square\]

This also works in reverse from the first coefficient.

Factorize \(25x^2-60x+36\).

We can factor out 5 from the middle coefficient and 25 from the first, and are left with \(x^2-12x+36\). Next, we can factor out 6 from the second coefficient and 36 from the final coefficient, and are left with \(x^2-2x+1=(x-1)^2\). Finally, we factor back in 5 to the first coefficient and 6 to the last one to obtain \((5x-6)^2\). \(_\square\)

This also works for unfactorable expressions. This method will not make unfactorable equations factorable; however, it will make the quadratic formula much easier to use. This is a little tougher to do because depending on which way you factor a number out, the formula changes.

**Case 1:** \(ax^2+bx+c\Rightarrow ax^2+\frac{b}{d}+\frac{c}{d^2}\). This changes the quadratic equation to

\[\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{2a}=\dfrac{-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}}{2a}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a\color{blue}{\textbf{d}}}.\]

Thus, once our answer is achieved, we must multiply our answer by the number \(\color{blue}{\textbf{d}}\) that we extracted at the start.

Factorize \(x^2+60x+2025\).

By factoring out \(15\) from the second coefficient and \(15^2=225\) from the final coefficient, we have

\[\begin{align} x^2+\frac{60}{15}x+\frac{2025}{225} &=x^2+4x+9 \\ &=\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\left(x-\frac{-4+\sqrt{16-36}}{2}\right)\\ &=\left(x-\big(-2+\sqrt{5}i\big)\right)\left(x-\big(-2-\sqrt{5}i\big)\right). \end{align}\]

Now, by multiplying back \(15,\) we have

\[\begin{align} x^2+60x+2025 &=\left(x-\big(-(15)2+(15)\sqrt{5}i\big)\right)\left(x-\big(-(15)2-(15)\sqrt{5}i\big)\right)\\ &=\left(x-\big(-30+15\sqrt{5}i\big)\right)\left(x-\big(-30-15\sqrt{5}i\big)\right).\ _\square \end{align}\]

**Case 2:** \(ax^2+bx+c\Rightarrow \frac{ax^2}{d^2}+\frac{b}{d}+c\). This changes the quadratic to

\[\dfrac{-\frac{b}{d}\pm \sqrt{\frac{b^2}{d^2}-\frac{4ac}{d^2}}}{\frac{2a}{d^2}}=\dfrac{d^2\left(-\frac{b}{d}\pm \frac{\sqrt{b^2-4ac}}{d}\right)}{2a}=\frac{\color{blue}{\textbf{d}}\big(-b\pm \sqrt{b^2-4ac}\big)}{2a}.\]

Thus, once our answer is achieved, we must divide our answer by the number \(\color{blue}{\textbf{d}}\) that we extracted at the start.

Now, we shall prove why this works.

Make the general expression \(ax^2+bx+c,\) which can be factored into \((dx+e)(fx+g).\) This means that \(a=df, b=dg+ef,\) and \(c=eg.\) The last step of our method requires us to multiply both of the second coefficients in our binomials by \(n\) \((n\) being the number that we factored out of \(b).\) So our expression becomes \(\big(dx+\frac{e}{n}\big)\big(fx+\frac{g}{n}\big),\) which means that \(a=df, \frac{b}{n}=\frac{dg+ef}{n}, \frac{c}{n^2}=\frac{eg}{n^2}\). This is why we factor out \(n\) and \(n^2\) from the second and third coefficients, respectively.

## Factorizing High-degree Polynomials with Large Numbers

This rule can actually apply to polynomials of any degree. Unfortunately, the higher the degree of the polynomial, the less convenient this becomes.

But say we have a polynomial with degree \(n\), which can be factored into \((a_1x+k_1)(a_2x+k_2)(a_3x+k_3)\cdots (a_nx+k_n)\). This does not necessarily have integer or real coefficients. Next, we assume that all coefficients after the second (inclusive) are divisible by \(r^t,\) where \(t\) represents the location of \(r\) relative to the second coefficient.

This means that we have \(\displaystyle\prod_{t=0}^n (a_tx+b_tr),\) which is equivalent to \(\displaystyle\sum_{t=0}^n \left(r^{n-t}\big(p_tx^t\big)\right).\) Upon expansion, we get

\[p_nx^n+r\big(p_{n-1}x^{n-1}\big)+r^2\big(p_{n-2}x^{n-2}\big)+r^3\big(p_{n-3}x^{n-3}\big)+\cdots+r^{n-1}\big(p_1x^1\big)+r^n(p_0).\]

As we can see, this method works whenever every coefficient after the first is divisible by \(r^t.\)

A useful method for seeing if this works is by prime factoring every coefficient after the second and looking for a common term. That has multiple powers throughout. Say we have \(x^4-3x^3-63x^2+27x+486\). First, we prime factor the numbers to get \(x^4-3x^3-3^2\cdot7x^2+3^3x+2\cdot3^5.\) As we can see, they share 3 in increasing powers. Therefore, we can eliminate 3 in increasing powers from each coefficient and are left with

\[\begin{align} x^4-x^3-7x^2+x+6&=x^3(x-1)-(7x+6)(x-1)\\ &=\left(x\big(x^2-1\big)-6(x+1)\right)(x-1)\\ &= \big(x^2-x-6\big)(x+1)(x-1)\\ &=(x-3)(x+2)(x+1)(x-1). \end{align}\]

Finally, we factor back in the three that we took out at the start and are left with \((x-9)(x+6)(x+3)(x-3).\)

**Cite as:**Factoring Polynomials With Large Coefficients: Factoring by Extraction.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/factoring-polynomials-with-large-coefficients/