Practice Factoring Trinomials with Two Variables

1. Teron and Abigail were each asked to factor the quadratic expression $$-y^2-4y+12$$. Their responses are shown below.

   Teron: $$-1(y-6)(y+2)$$
   

   Abigail: $$(-y+6)(y-2)$$
   

   Which student wrote an expression that is equivalent to $$-y^2-4y+12$$?

   Choose 1 answer:

   1. Only Teron
   2. Only Abigail
   3. Both Teron and Abigail

   4. Neither Teron nor Abigail

2. The polynomial expressions ‍$$12y^3-5y^2-3y$$ and $$5(3y+1)(2y-3)$$ share a common binomial factor.

   What binomial factor do they share?

3. The quadratic expression ‍$$x^2+Mx+30$$ can be factored as ‍$$(x+a)(x+b)$$, where ‍$$M$$, ‍‍$$a$$, ‍and ‍$$b$$ are integers.

   Jusna wrote that ‍$$M$$ could be equal to -13.

   Ethan wrote that $$M$$ could be equal to ‍11.
   

   Which student is correct ?
   

   Choose 1 answer:

   1. Only Jusna
   2. Only Ethan
   3. Both Jusna and Ethan

   4. Neither Jusna nor Ethan

4. The rectangle below has an area of $$12ax-28ay+15bx-35by$$ square meters and a width of $$4a+5b$$ meters.

   What expression represents the length of the rectangle?

   
   

Source: Khan Academy, https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-factoring-polynomials-quadratic-forms/e/factoring_polynomials_with_two_variables
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. 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 $$-y^2-4y+12$$.

2. 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 ‍$${3y+1}$$:
   

   $$\begin{aligned} 12y^3-5y^2-3y&=y({3y+1})(4y-3)\\\\
   5(3y+1)(2y-3)&=5({3y+1})(2y-3)
   

   \end{aligned}$$

3. 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.

4. 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: