Equation Lab #16 – Quadratic Equations: Factoring (1)

Yesterday’s lesson introduced you to quadratic equations. But you would only know how to solve quadratic equations without the linear term. Now what if there is a linear term? Then you can’t use the square roots method. Instead, you’ll have to use a more advanced method.

For the next three days, I will be covering on how to solve quadratic equations by factoring. Since we haven’t covered how to multiply or factor polynomials, I will go over that as well for each lesson. The first one will be on how to factor $x^{2}+bx+c$, the second one will be on how to factor $ax^{2}+bx+c$, and the third one will be on how to factor special cases. Once all three lessons are covered, we will move onto the next method.

Solving by Factoring – Basic Functions:

Before I even go over how to factor quadratic trinomials, I would like to cover an important property of solving equations. This is otherwise known as the zero-product property. As previously mentioned, the solutions to a quadratic equation are the x-intercepts, where $y=0$. That means, the entire trinomial is equal to $0$ when we evaluate the variable of $x$.

Given that any number times $0$ always equal $0$, the quadratic trinomial in factored form will have a product of $0$. That means either linear binomial equals $0$.

So let’s say that $(ax+b)(cx+d)=0$. Either $ax+b=0$, or $cx+d=0$. In each case, you will have to solve for $x$. This is when you will get two different solutions.

So the algebraic property of the lesson is:

• Zero-Product Property:
  • $(a+b)(c+d)=0$
  • $a+b=0$ or $c+d=0$

So let’s go over how to factor trinomials. When $a=1$, what you need to do is to find two factors of $c$ that add up to $b$. A formula for solving by factoring is:

• $x^{2}+bx+c=(x+c_{1})(x+c_{2})$
• $b=c_{1}+c_{2}; c=c_{1}\cdot c_{2}$

In the case where both $b$ and $c$ are positive, both $c_{1}$ and $c_{2}$ are positive. In the case where $c$ is positive, but $b$ is negative, both $c_{1}$ and $c_{2}$ are negative. In any case where $c$ is negative, one of the factors is positive, while the other is negative. $b$ will only be the difference between $c_{1}$ and $c_{2}$.

With that out of the way, let’s start with our first example. The equation is $w^{2}-14w+40=0$. What two factors of $40$ add up to $14$? For a recall, here are the factors of $40$:

• $1, 2, 4, 5, 8, 10, 20, 40$

As you know, $1\cdot40=40$, $2\cdot20=40$, $4\cdot10=40$, and $8\cdot5=40$. But if we changed the operations to addition, but kept the numbers, the sums will not be the same. The sum that matches $b$, which in this case $14$, is what we’re looking for.

• $1+40=41$
• $2+20=22$
• $4+10=14$
• $5+8=13$

Since $4+10=14$, the factors of $40$ to use are $4$ and $10$. But $14$ is negative. Therefore, $w^{2}-14w+40$ is the same as $(w-4)(w-10)$.

We now have the equation $(w-4)(w-10)=0$. This is when we can start using the zero-product property. If $(w-4)(w-10)=0$, then either $w-4=0$, or $w-10=0$. You’ll have to solve both equations independently. For $w-4=0$, you’ll have to add $4$ to both sides to solve the equation. Your result is $w=4$. For $w-10=0$, you’ll have to add $10$ to both sides to solve the equation. The result is $w=10$. Therefore, $w=4$ or $w=10$.

$w^{2}-14w+40$
$b=-14, c=40$

Factors of $40$:
- $1, 2, 4, 5, 8, 10, 20, 40$

- $1\cdot 40=40; 1+40=41$   X
- $2\cdot20=40; 2+20=22$   X
- $4\cdot10=40; 4+10=14$   ✓
- $5\cdot 8=40; 5+8=13$     X

$(w-4)(w-10)$

1. $w^{2}-14w+40=0$
2. $(w-4)(w-10)=0$
3. $w-4=0, w-10=0$
4. $w=4, w=10$

Now let’s check our work to see if either $w=4$ or $w=10$ is correct. When $w=4$, $w^{2}=16$ and $-14w=-56$. $40+16=56$, but $56-56=0$. So when $w=4$, $w^{2}-14w+40=0$. What about $w=10$ In this case, $w^{2}=100$ and $-14w=-140$. $100+40=140$, but $140-140=0$. So when $w=10$, $w^{2}-14w+40=0$. This means that both $w=4$ and $w=10$ are correct.

Let’s do another example. We want to solve $l^{2}+3l-28=0$. What two factors of $28$ have a difference of $3$? For a recall, here are the factors of $28$:

• $1, 2, 4, 7, 14, 28$

We get that $1\cdot 28$, $2\cdot 14$, and $4\cdot 7$ all equal $28$. Since $28$ is negative, we are looking for differences, not sums.

• $28-1=27$
• $14-2=12$
• $7-4=3$

Since $7-4=3$, the factors of $28$ to use are $4$ and $7$. And since $3$ is positive, that should make $7$ positive and $4$ negative. Therefore, $l^{2}+3l-28$ is the same as $(l+7)(l-4)$.

Our equation to solve is $(l+7)(l-4)=0$. Using the zero-product property, either $l+7=0$ or $l-4=0$. For $l+7=0$, you can solve it by subtracting $7$ from both sides. The result is $l=-7$. For $l-4=0$, you can solve it by adding $4$ to both sides. The result is $l=4$. Therefore, $l=-7$ or $l=4$.

$l^{2}+3l-28$
$b=3; c=-28$

Factors of $28$:
- $1, 2, 4, 7, 14, 28$

- $1\cdot28=28; 28-1=27$   X
- $2\cdot 14=28; 14-2=12$   X
- $4\cdot 7=28; 7-4=3$      ✓

$(l+7)(l-4)$

1. $l^{2}+3l-28=0$
2. $(l+7)(l-4)=0$
3. $l+7=0,l-4=0$
4. $l=-7,l=4$

Now let’s check our work to see if either $l=-7$ or $l=4$ is correct. When $l=-7$, $l^{2}=49$ and $3l=-21$. $-28-21=-49$, but $49=49=0$. So when $l=-7$, $l^{2}+3l-28=0$. When $l=4$, $l^{2}=16$ and $3l=12$. $16+12=28$, but $28-28=0$. So when $l=4$, $l^{2}+3l-28=0$. This means that both $l=-7$ and $l=4$ are correct.

Word Problem #14:

A new room in a house has an area of 216 square feet. Compared to the previous room built, which was a perfect square, the length is 4 feet longer than the other room’s length, and the width is 2 feet shorter than the other room’s length. What are the dimensions of the new room? How much bigger is the new room?

This one will take multiple steps to solve. So we are given that the area of the new room is $216$ square feet. The area of the old room is unknown, but since it’s a perfect square, we’ll use $s^{2}$ for the area of the old room. As for the new room, the width is $2$ feet shorter than the length of each wall in the old room, but the length is $4$ feet longer than the length of every room. We use $s-2$ to denote the width since it’s two feet shorter, and $s+4$ to denote the length since it’s four feet longer. The equation to solve is $(s-2)(s+4)=216$.

The very first step is to multiply $s-2$ to $s+4$. Using the FOIL method, the products are $s^{2}$, $4s$, $-2s$, and $-8$. The trinomial is $s^{2}+2s-8$. We now have the equation $s^{2}+2s-8=216$. Since you can’t use the zero-product property here, you’ll have to subtract $216$ from both sides. You will get $s^{2}+2s-224=0$. Now we’ll have to factor the new trinomial, which is $s^{2}+2s-224$. Considering the factors of $224$, they are:

• $1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 112, 224$

Which of these two factors have a difference of $2$? Let’s take a look.

• $224-1=223$
• $112-2=110$
• $56-4=52$
• $32-7=25$
• $28-8=20$
• $16-14=2$

It seems that $14$ and $16$ have a difference of $2$, and $2$ is positive. For these two reasons, $s^{2}+2s-224$ is the same as $(s+16)(s-14)$.

Now let’s solve, using the zero-product property. If $s^{2}+2s-224=0$, then $(s+16)(s-14)=0$. This means that either $s+16=0$ or $s-14=0$. Solving each equation, you should get $s=-16$ and $s=14$. But since we are solving a word problem about measurements, we must disregard $s=-16$ and take $s=14$.

While the equation has been solved, neither question have been answered. The first one is to calculate the dimensions of the new room. If $s=14$, then $s-2=12$ and $s+4=18$. So the room is 18 feet long and 12 feet wide. As for the second question, since the comparisons are based on the old room, it seems that the old room is 14 feet long and 14 feet wide. The area should be $196$. Now we must subtract both areas. $216-196=20$. Therefore, the new room is 20 square feet larger than the old room.

$s^{2}+2s-224$
$b=2; c=-224$

Factors of $224$:
- $1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 112, 224$

- $1\cdot 224=224; 224-1=223$  X
- $2\cdot112=224; 112-2=110$  X
- $4\cdot 56=224; 56-4=52$     X
- $7\cdot 32=224; 32-7=25$     X
- $8\cdot28=224; 28-8=20$     X
- $14\cdot16=224; 16-14=2$    ✓

$(s+16)(s-14)$

 1. New room is 216 square feet.
 2. Dimensions of new room is 4 feet
    longer and 2 feet shorter than
    older room's dimensions.
 3. $(s-2)(s+4)=216$
 4. $s^{2}+2s-8=216$
 5. $s^{2}+2s-224=0$
 6. $(s+16)(s-14)=0$
 7. $s+16=0, s-14=0$
 8. $s=-16, s=14$
 9. $s=14; s-2=12, s+4=18$, $s^{2}=196$
10. $216-196=20$
11. Dimensions are 12 feet and 18 feet.
    New room is 20 square feet larger.

Quiz #16:

Now that the lesson is over, let’s see if you can complete this quiz. Like always, the correct solutions and the correct variables must be used. But for this quiz, I will take any answer for as long as it fits the solution. So if the answer is $a=3,-3$, you can use “$3$“, “$-3$“, “$3,-3“$, “$-\mathrm{3,3}$“, “$a=3$“, “$a=-3$“, “$a=3,-3$“, or “$a=-\mathrm{3,3}$“.