Factoring polynomials using grouping

The idea of grouping

In this lesson we’ll look at factoring a polynomial using a method called grouping.

When you have a polynomial, sometimes you can use factoring by grouping to help you get the factored parts.

It means you need to look for terms in the polynomial that have values and terms in common and then group those parts together.

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How to use grouping to factor polynomials

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Three more grouping examples

Example

Factor by grouping.

$11z+11qr+pyz+pyqr$

Since we’ve been asked to use grouping to factor the polynomial, we need to look for terms in the polynomial that have values in common and group them together. Since the first two terms have an $11$ in common, and the last two terms have a $py$ in common, we’ll group the first two terms together separately from the last two terms.

$11z+11qr+pyz+pyqr$

$(11z+11qr)+(pyz+pyqr)$

With our terms grouped, we need to look for the greatest common factor in each group. In this case, those are the values we identified earlier that we used to group the terms together. Factoring these out of each group separately, we get

$11(z+qr)+py(z+qr)$

We always want to factor as completely as possible, which means we need to factor a $z+qr$ out of each term above. Pulling $z+qr$ out of each term leaves us with

$(11+py)(z+qr)$

This is the correct solution, but it can also be written as $(z+qr)(11+py)$ or even $(qr+z)(11+py)$.

Remember, there are usually multiple ways to group our terms before we factor. We could have also solved it this way:

$11z+11qr+pyz+pyqr$

$11qr+pyqr+11z+pyz$

$(11qr+pyqr)+(11z+pyz)$

$qr(11+py)+z(11+py)$

$(qr+z)(11+py)$

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Let’s do one more example.

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With our terms grouped, we need to look for the greatest common factor in each group.

Example

Factor by grouping.

$pqx^2-psx+qrx-rs$

Since we’ve been asked to use grouping to factor the polynomial, we need to look for terms in the polynomial that have values in common and group them together. Since the first two terms have a $px$ in common, and the last two terms have an $r$ in common, we’ll group the first two terms together separately from the last two terms.

$pqx^2-psx+qrx-rs$

$(pqx^2-psx)+(qrx-rs)$

With our terms grouped, we need to look for the greatest common factor in each group. In this case, those are the values we identified earlier that we used to group the terms together. Factoring these out of each group separately, we get

$px(qx-s)+r(qx-s)$

We always want to factor as completely as possible, which means we need to factor a $qx-s$ out of each term above. Pulling $qx-s$ out of each term leaves us with

$(px+r)(qx-s)$

As we mentioned before there are multiple ways of writing this, such as $(-s+qx)(r+px)$, and it all depends on how you choose to group the factors.

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We can also use grouping to factor quadratics. We already know how to factor quadratics of the form

$ax^2+bx+c$

by looking at the factors of $a$ and $c$ and trying to figure out which combination of factors can be used to make $b$. But that’s not the only way to factor quadratics. We can also use the grouping method.

First, let’s review the pieces of a quadratic expression. Any quadratic can be written as $ax^2+bx+c$, where $a$, $b$ and $c$ are real numbers and $a \neq 0$. Factoring means you can break the quadratic into parts like

$ax^2+bx+c=(px+r)(qx+s)$

where $p$, $q$, $r$, and $s$ are also real numbers.

You can use factoring by grouping to help you get these parts. It means you need to look for terms in the polynomial that have values that are the same and then group those parts together.

Let’s look at bit closer at what’s going on. Let’s multiply $(px+r)(qx+s)$ to look at the parts.

$(px+r)(qx+s)=pqx^2+psx+qrx+rs$

$(px+r)(qx+s)=pqx^2+(ps+qr)x+rs$

If we match up this right side to the standard form of a quadratic, we get

$ax^2+bx+c=pqx^2+(ps+qr)x+rs$

Then, if we match up the coefficients on $x^2$ and $x$, and the constant, we get $a=pq$, $b=ps+qr$ and $c=rs$. Also $a \cdot c = ps \cdot qr$.

Now look at the factor pairs of $a\cdot c$ and the sum of the factor pairs for the $b$ term.

Let’s look at an example with numbers to see how this works out.

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Example

Factor the quadratic.

$11x^2+13x+2$

We want to factor this so it looks like $(px+r)(qx+s)$.

Let’s look at $a\cdot c$ and the sum of the factor pairs for the $b$ term.

In this case $a=11$, $c=2$ and $b=13$, so we need to find the factor pairs of $a\cdot c=11\cdot 2=22$ that add to $13$. The factor pairs of $22$ are $1\cdot 22$ and $2\cdot11$. Which of these factor pairs add to $13$?

$1+22=23$

and

$2+11=13$

So we need the factor pair $2,\ 11$. Let’s rewrite $11x^2+13x+2$ as $11x^2+2x+11x+2$.

Now, since we want to use grouping to factor the polynomial, we need to look for terms in the polynomial that have values in common and group them together. Since the first two terms have an $x$ in common, we’ll group the first two terms together separately from the last two terms.

$(11x^2+2x)+(11x+2)$

Let’s factor out the $x$ from the first term.

$x(11x+2)+(11x+2)$

Remember we can write this as

$x(11x+2)+1(11x+2)$

We always want to factor as completely as possible, which means we need to factor an $11x+2$ out of each term above. Pulling $11x+2$ out of each term leaves us with

$(11x+2)(x+1)$

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If you’re struggling with factoring a quadratic, factoring by grouping can give you a nice procedure to follow.

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