Math Trailhead

\(25\) and \(30\) have a common factor of \(5\text{.}\)

\(x^{4}\) and \(x^{3}\) have a common factor of \(x^{3}\)

\(18\) and \(6\) have a common factor of \(6\text{.}\)

\(x^{2}\) and \(x^{1}\) have a common factor of \(x^{1}\)

So, \({18x^{2}-6x} \rightarrow {6x^{1}\mathopen{}\left(3x-1\right)}\text{.}\)

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

\(s(10s^2 - 9s - 7)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(10\cdot-7=-70\) and a sum of \(-9\text{.}\) Those numbers are \(5\) and \(-14\text{.}\)

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

\(s(10s^2 + 5s - 14s - 7)\)

Now we can factor by grouping

\(s(5s( 2s + 1 ) - 7( 2s + 1 ))\)

\({s\mathopen{}\left(2s+1\right)\mathopen{}\left(5s-7\right)}\)

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

\(3w(10w^2+39w + 35)\)

Now we can use the AC-method to factor the remaining trinomial. We are looking for two numbers with a product of \(10\cdot35=350\) and a sum of \(39\text{.}\) Those numbers are \(25\) and \(14\text{.}\)

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

\(3w(10w^2 + 25w + 14w + 35)\)

Now we can factor by grouping

\(3w(5w( 2w + 5 ) + 7( 2w + 5 ))\)

\({3w\mathopen{}\left(2w+5\right)\mathopen{}\left(5w+7\right)}\)

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

\(3({t^{3}+5t^{2}+3t+15})\)

Then we can factor the remaining expression by grouping.

\(3(t^2(t+5) + 3(t+5))\)

\({3\mathopen{}\left(t^{2}+3\right)\mathopen{}\left(t+5\right)}\)

\(2({y+8})\) and \(7({y+8})\) have a common factor of \({y+8}\text{.}\)

Think about it. \({y+8}\) will be the same value in each term, no matter what value \(y\) represents.

Each term will be a multiple of \({y+8}\text{,}\) so it is a common factor.

So, \({2x\mathopen{}\left(y+8\right)-7\mathopen{}\left(y+8\right)} \rightarrow {\left(y+8\right)\mathopen{}\left(2x-7\right)}\text{.}\)

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

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

\((15 A B + 21 A) + (25 B + 35)\)

Factor each group separately:

\((15 A B + 21 A) \rightarrow 3 A ( 5 B + 7 )\)

\((25 B + 35) \rightarrow 5 ( 5 B + 7 )\)

Each term is a multiple of \(5 B + 7\text{,}\) so it is a common factor.

So, \({15AB+21A+25B+35} \rightarrow 3 A ( 5 B + 7 ) + 5 ( 5 B + 7 )\)

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

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

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

\(-9 A B + 21 A - 6 B + 14\)

\((-9 A B + 21 A) + (-6 B + 14)\)

Factor each group separately:

\((-9 A B + 21 A) \rightarrow 3 A ( -3 B + 7 )\)

\((-6 B + 14) \rightarrow 2 ( -3 B + 7 )\)

Each term is a multiple of \(-3 B + 7\text{,}\) so it is a common factor.

So, \({-9AB+21A-6B+14} \rightarrow 3 A ( -3 B + 7 ) + 2 ( -3 B + 7 )\)

and \(3 A ( -3 B + 7 ) + 2 ( -3 B + 7 ) \rightarrow {\left(-3B+7\right)\mathopen{}\left(3A+2\right)}\)

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

\(\begin{aligned} {8x^{2}+72xy+3xy+27y^{2}}\amp =8x(x+9y)+3y(x+9y) \\ \amp ={\left(8x+3y\right)\mathopen{}\left(x+9y\right)} \end{aligned}\)

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.

\(\begin{aligned} {36t^{5}+72t^{4}-36t^{4}-72t^{3}+45t^{3}+90t^{2}}\amp =({36t^{5}+72t^{4}})+({-36t^{4}-72t^{3}})+({45t^{3}+90t^{2}})\\ \amp ={36t^{4}\mathopen{}\left(t+2\right)-36t^{3}\mathopen{}\left(t+2\right)+45t^{2}\mathopen{}\left(t+2\right)}\\ \amp ={9t^{2}\mathopen{}\left(t+2\right)\mathopen{}\left(4t^{2}-4t+5\right)} \end{aligned}\)