Math Trailhead

\(21\) and \(9\) have a common factor of \(3\text{.}\)

\(x^{5}\) and \(x^{4}\) have a common factor of \(x^{4}\)

\(20\) and \(4\) have a common factor of \(4\text{.}\)

\(x^{5}\) and \(x^{4}\) have a common factor of \(x^{4}\)

So, \({20x^{5}-4x^{4}} \rightarrow {4x^{4}\mathopen{}\left(5x-1\right)}\text{.}\)

This problem has a greatest common factor (GCF) of \(m\text{.}\) If we factor the GCF out, we have

\(m(12m^2 - 19m + 5)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(12\cdot5=60\) and a sum of \(-19\text{.}\) Those numbers are \(-15\) and \(-4\text{.}\)

We can use those two numbers to rewrite the middle term as two terms:

\(m(12m^2 - 15m - 4m + 5)\)

Now we can factor by grouping

\(m(3m( 4m - 5 ) - 1( 4m - 5 ))\)

\({m\mathopen{}\left(4m-5\right)\mathopen{}\left(3m-1\right)}\)

This problem has a greatest common factor (GCF) of \(3p\text{.}\) If we factor the GCF out, we have

\(3p(6p^2 - 5p + 1)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(6\cdot1=6\) and a sum of \(-5\text{.}\) Those numbers are \(-3\) and \(-2\text{.}\)

We can use those two numbers to rewrite the middle term as two terms:

\(3p(6p^2 - 3p - 2p + 1)\)

Now we can factor by grouping

\(3p(3p( 2p - 1 ) - 1( 2p - 1 ))\)

\({3p\mathopen{}\left(2p-1\right)\mathopen{}\left(3p-1\right)}\)

There is a greatest common factor of \(3\text{.}\) Start by factoring that out.

\(3({r^{3}-8r^{2}-6r+48})\)

Then we can factor the remaining expression by grouping.

\(3(r^2(r - 8) - 6(r - 8))\)

\({3\mathopen{}\left(r^{2}-6\right)\mathopen{}\left(r-8\right)}\)

\(6({y+6})\) and \(7({y+6})\) have a common factor of \({y+6}\text{.}\)

Think about it. \({y+6}\) will be the same value in each term, no matter what value \(y\) represents.

Each term will be a multiple of \({y+6}\text{,}\) so it is a common factor.

So, \({6x\mathopen{}\left(y+6\right)-7\mathopen{}\left(y+6\right)} \rightarrow {\left(y+6\right)\mathopen{}\left(6x-7\right)}\text{.}\)

Begin by grouping the first two terms and last two terms:

We are allowed to do this by the associative property of addition.

\((14 A B + 49 A) + (4 B + 14)\)

Factor each group separately:

\((14 A B + 49 A) \rightarrow 7 A ( 2 B + 7 )\)

\((4 B + 14) \rightarrow 2 ( 2 B + 7 )\)

Each term is a multiple of \(2 B + 7\text{,}\) so it is a common factor.

So, \({14AB+49A+4B+14} \rightarrow 7 A ( 2 B + 7 ) + 2 ( 2 B + 7 )\)

and \(7 A ( 2 B + 7 ) + 2 ( 2 B + 7 ) \rightarrow {\left(2B+7\right)\mathopen{}\left(7A+2\right)}\)

Begin by grouping the first two terms and last two terms:

You must be careful, subtraction is not associative like addition is.

\(-35 A B + 21 A - 30 B + 18\)

\((-35 A B + 21 A) + (-30 B + 18)\)

Factor each group separately:

\((-35 A B + 21 A) \rightarrow 7 A ( -5 B + 3 )\)

\((-30 B + 18) \rightarrow 6 ( -5 B + 3 )\)

Each term is a multiple of \(-5 B + 3\text{,}\) so it is a common factor.

So, \({-35AB+21A-30B+18} \rightarrow 7 A ( -5 B + 3 ) + 6 ( -5 B + 3 )\)

and \(7 A ( -5 B + 3 ) + 6 ( -5 B + 3 ) \rightarrow {\left(-5B+3\right)\mathopen{}\left(7A+6\right)}\)

There are a few ways to approach this type of problem- we will demonstrate the factor by grouping method, but it can be done in other ways

\(\begin{aligned} {5x^{2}+15xy+6xy+18y^{2}}\amp =5x(x+3y)+6y(x+3y) \\ \amp ={\left(5x+6y\right)\mathopen{}\left(x+3y\right)} \end{aligned}\)

Note that this answer can be checked by using the FOIL (First Outside Inside Last) technique (exercise).

Let’s follow the hint and begin by factoring each pair of terms first- this should help to highlight the pattern.

\(\begin{aligned} {32r^{5}+64r^{4}-24r^{4}-48r^{3}+40r^{3}+80r^{2}}\amp =({32r^{5}+64r^{4}})+({-24r^{4}-48r^{3}})+({40r^{3}+80r^{2}})\\ \amp ={32r^{4}\mathopen{}\left(r+2\right)-24r^{3}\mathopen{}\left(r+2\right)+40r^{2}\mathopen{}\left(r+2\right)}\\ \amp ={8r^{2}\mathopen{}\left(r+2\right)\mathopen{}\left(4r^{2}-3r+5\right)} \end{aligned}\)