Practice Factoring Trinomials with Two Variables

Let's factor the quadratic expression and compare our answer to what Teron and Abigail wrote.

Each term in the quadratic expression shares the common factor -1, so we can begin by factoring this out.

\(-y^2-4y+12=-1(y^2+4y-12)\)

We can now focus on factoring the quadratic expression \(y^2+4y-12\).

In order to factor \(y^2{+4}y{-12}\) as ‍\((y+a)(y+b)\), we need to find two integers ‍\(a\) and ‍\(b\) and ‍whose sum is 4 and whose product is -12. We can find them by first considering all pairs of integers whose product is -12, and then calculating the sums of those pairs.

It turns out that \(a=6\) and ‍\(b=-2\) satisfy these conditions, since ‍\(6-2=4\) and \((6)(-2)={-12}\).

We now know that \(y^2+4y-12=(y+6)(y-2)\).

When we put our two factorizations together, we can conclude that \(-y^2-4y+12=-1(y+6)(y-2)\).

Teron wrote \(-1(y-6)(y+2)\). His expression is not equivalent to ‍\(-y^2-4y+12\) because he mixed up his signs. Teron's expression is equivalent to \(-y^2+4y+12\).

Abigail wrote \((-y+6)(y-2)\). Her expression is not equivalent to ‍ \(-y^2-4y+12\) because she also mixed up her signs. Abigail's expression is equivalent to \(-y^2+8y-12\).

Neither Teron nor Abigail wrote an expression that is equivalent to \(-y^2-4y+12\).

Let's factor the polynomial expression \(12y^3-5y^2-3y\) and then compare it with the binomial factors of \(5(3y+1)(2y-3)\).

Each term in this polynomial expression shares the common monomial factor ‍\(y\), so we can factor this out.

\(12y^3-5y^2-3y=y(12y^2-5y-3)\)

We can now focus on factoring the quadratic expression \({12}y^2{-5}y{-3}\) by grouping.

Our first step is to find two integers ‍\(a\) and ‍\(b\) such that:

\( a+b={-5} \\ \quad ab=(12)(-3)=-36 \)

We can find them by first considering all pairs of integers whose product is ‍-36, and then calculating the sums of those pairs.

It turns out that \(a=-9\) and ‍\(b=4\) satisfy these conditions, since ‍\(-9+4={-5}\) and \((-9)(4)=-36\).

We can use these two integer values to decompose the \(y\) term, and continue our factorization by grouping:

\(\begin{aligned}&\phantom{{}={}}{12}y^2{-5}y{-3}\\\\\\
&={12}y^2-9y+4y{-3}&\text{Decompose the }y\text{ term}\\\\
&=({12}y^2-9y)+(4y{-3})&\text{Group into binomial pairs}\\\\
&=3y(4y-3)+1(4y-3)&\text{Factor each binomial pair}\end{aligned}\)

Now we can factor ‍\(4y-3\) out of the entire expression:

When we put our two factorizations together, we see that \(12y^3-5y^2-3y=y(3y+1)(4y-3)\).

In conclusion, both of the polynomial expressions share the binomial factor ‍\({3y+1}\):

We can factor a quadratic expression as the product of two binomials by using the sum/product pattern:

\(x^2 +{(a+b)}\ x + {ab}=(x+a)(x+b)\)

Since \(x^2+Mx+30\) can also be factored as \((x+a)(x+b)\), we know that the following equality holds:

\(x^2{+M}x{+30}=x^2 +{(a+b)}\ x + {ab}\)

Therefore, we know that \(a+b={M}\) and \(ab={30}\).

The pairs of integers whose product is 30 are listed below:

• \((+1\) and \(+30)\) ‍or ‍\((-1\) and \(-30)\)
• \((+2\) and ‍$+15)$ or ‍\((-2\) and \(-15)\)
• \((+3\) and ‍\(+10)\) or ‍\((-3\) and \(-10)\)

• \((+5\) and ‍\(+6)\) or ‍\((-5\) and \(-6)\)

Any of these eight pairs of integers could be the values for \(a\) and ‍\(b\). Therefore, there are only eight possible values for \(a+b={M}\):

\(\begin{aligned}{M}&=(\pm 1)+(\pm 30)=\pm31\\\\
{M}&=(\pm 2)+(\pm 15)=\pm 17\\\\
{M}&=(\pm 3)+(\pm 10)=\pm 13\\\\
{M}&=(\pm 5)+(\pm 6)=\pm 11
\end{aligned}\)

Jusna wrote that \(M\) could be equal to ‍-13, so Jusna is correct. If \(M=-13\), the given quadratic expression could be factored as:

\(x^2-13x+30=(x-3)(x-10)\)

Ethan wrote that \(M\) could be equal to ‍11, so Ethan is also correct. If \(M=11\), the given quadratic expression could be factored as:

\(x^2+11x+30=(x+5)(x+6)\)

Both Jusna and Ethan are correct.

The area of any rectangle is the product of its length \((L)\)
and width \((W)\):

\(Area =(L)(W)\)

This rectangle has an area of \(12ax-28ay+15bx-35by\) and a length of \(4a+5b\):

\(12ax-28ay+15bx-35by =(L)(4a+5b)\)

Let's find \(L\) by factoring the area expression!

We can factor this expression by grouping:

\(\begin{aligned}&\phantom{{}={}}12ax-28ay+15bx-35by\\\\
&=(12ax-28ay)+(15bx-35by)&\text{Group into binomial pairs}\\\\
&=4a(3x-7y)+5b(3x-7y)&\text{Factor each binomial pair}\end{aligned}\)

It turns out that \(3x-7y\) is a factor shared by each pair of terms! Let's factor it out:

Since the width of the rectangle is \(4a+5b\) meters, we now know that the length of the rectangle must be \(3x-7y\) meters.

Notice that the four terms in the area expression \(12ax-28ay+15bx-35by\) correspond to areas of four smaller rectangles that make up the large rectangle. This should make sense, as the whole is the sum of its parts!

In conclusion, \(\text{Length }=3x-7y\) meters: