Equation Lab #17 – Quadratic Equations: Factoring (2)

As the solving by factoring lessons continue, we are onto the next level, which is solving quadratic equations with a leading coefficient other than $1$. We previously learned how to solve quadratic equations by factoring when the leading coefficient is $1$, but now we’re going to the next level.

Solving by Factoring – Advanced Functions:

So we’re now going into how to solve equations of $ax^{2}+bx+c=0$. If $ax^{2}+bx+c$ is factorable, then we should get the following formulas:

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

If $a$, $b$, and $c$ have the same sign, then $a_{1}$, $a_{2}$, $c_{1}$, and $c_{2}$ are all positive. If $a$ and $c$ have the same sign, but $b$ has the opposite sign, then $a_{1}$ and $a_{2}$ are positive while $c_{1}$ and $c_{2}$ are negative. In both cases, if $a$ is negative, you can factor out the negative sign from the whole trinomial.

If $a$ and $c$ do not have the same sign, then there’s going to be a mix in positive and negative numbers. If $a$ is negative and $c$ is positive, then both $c_{1}$ and $c_{2}$ are positive while $a_{1}$ and $a_{2}$ have different signs. If $a$ is positive and $c$ is negative, then both $a_{1}$ and $a_{2}$ are positive while $c_{1}$ and $c_{2}$ have different signs.

In today’s lesson, we’ll use the cases where $a$ is a positive integer.

Let’s start with the first equation. $3q^{2}+18q+24=0$. This one seems easier because all three terms have a common factor, which is $3$. If you divide all terms by $3$, you will get $q^{2}+6q+8=0$. We can solve this equation like how we solved previous equations. Since $8$ is a four-factored number, our only sums generated by the factors are $9$ ($1+8$) and $6$ ($2+4$). The linear term’s coefficient is $6$, so $q^{2}+6q+8$ breaks down into $(q+4)(q+2)$, which in return, $3q^{2}+18q+24$ breaks down into $3(q+4)(q+2)$.

Now let’s use the zero-product property in action. If $(q+4)(q+2)=0$, then either $q+4=0$ or $q+2=0$. Solving both equations, we get $q=-4$ and $q=-2$.

$3q^{2}+18q+24$
$a=3; b=18; c=24$

$3(q^{2}+6q+8)$

Factors of $8$:
- $1,2,4,8$

- $1\cdot8=8; 1+8=9$ X
- $2\cdot 4=8; 2+4=6$ ✓

$3(q+4)(q+2)$

1. $3q^{2}+18q+24=0$
2. $3(q+4)(q+2)=0$
3. $q+4=0,q+2=0$
4. $q=-4, q=-2$

Time for the next problem, $6n^{2}-23n+20=0$. This time, there are no common factors among all three terms. This is where the real strategy has to be used. We need to find the factors of both $6$ and $20$ that multiply into the two numbers that add into $23$. For a reminder, here are the factors:

• $6$: $1,2,3,6$
• $20$: $1,2,4,5,10,20$

What you’ll have to do is to cross the factors of $6$ and $20$ over. We know that $1$ and $6$ multiply into $6$. Let’s try that.

• $1\cdot 1+6\cdot 20=121$
• $1\cdot 2+6\cdot 10=62$
• $1\cdot 4+6\cdot 5=34$
• $1\cdot 5+6\cdot 4=29$
• $1\cdot10+6\cdot 2=22$
• $1\cdot 20+6\cdot 1=26$

Oh dear! None of the sums generated are $23$. This means that $1$ and $6$ are not the factors of $a$ in the factored form. Let’s try $2$ and $3$ instead.

• $2\cdot1+3\cdot 20=62$
• $2\cdot 2+3\cdot 10=34$
• $2\cdot 4+3\cdot 5=23$
• $2\cdot 5+3\cdot 4=22$
• $2\cdot 10+3\cdot2=26$
• $2\cdot 20+3\cdot 1=43$

It looks like when the factors of $20$ are $4$ and $5$, we get a sum of $23$. Now what two binomials are correct? Is it $2n-4$ and $3n-5$, or $2n-5$ and $3n-4$? If you remember, the FOIL method does not multiply two numbers within the same binomial. So if we got $2n-4$ and $3n-5$, the FOIL method would calculate $2\cdot-5+3\cdot-4$. But if we got $2n-5$ and $3n-4$, we would calculate $2\cdot-4+3\cdot -5$, which is $-23$. So, $6n^{2}-23n+20$ is the same as $(2n-5)(3n-4)$.

Now we can use the zero-product property. If $6n^{2}-23n+20=0$, then $(2n-5)(3n-4)=0$. That means that either $2n-5=0$ or $3n-4=0$. For $2n-5=0$, if we add $5$ to both sides, we get $2n=5$. Divide both sides by $2$, we get $n=\frac{5}{2}$. For $3n-4=0$, if we add $4$ to both sides, we get $3n=4$. Divide both sides by $3$, we get $n=\frac{4}{3}$. Our solutions are $n=\frac{5}{2}$ and $n=\frac{4}{3}$.

$6n^{2}-23n+20$
$a=6; b=-23; c=20$

Factors of $6$:
- $1, 2, 3, 6$

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

- $1\cdot 1+6\cdot 20=121$  X
- $2\cdot 1+3\cdot 20=62$   X
- $1\cdot 2+6\cdot 10=62$   X
- $2\cdot 2+3\cdot 10=34$   X
- $1\cdot 4+6\cdot 5=34$    X
- $2\cdot 4+3\cdot 5=23$    ✓
- $1\cdot 5+6\cdot 4=29$    X
- $2\cdot 5+3\cdot4=22$    X
- $1\cdot 10+6\cdot 2=22$   X
- $2\cdot 10+3\cdot 2=26$   X
- $1\cdot 20+6\cdot 1=26$   X
- $2\cdot 20+3\cdot 1=43$   X

$(2n-5)(3n-4)$

1. $6n^{2}-23n+20=0$
2. $(2n-5)(3n-4)=0$
3. $2n-5=0, 3n-4=0$
4. $2n=5, 3n=4$
5. $n=\frac{5}{2}, n=\frac{4}{3}$

Word Problem #15:

The city limit is 8,000 meters away from home. A car, moving at 11 meters per second, began accelerating at 4 meters per squared second. How long does it take for it to be 8,105 meters away from home?

First, let’s set up the equation. This looks like a Physics equation. In Physics, the horizontal position formula is $x=x_{0}+v_{x}t+\frac{1}{2}a_{x}t^{2}$. We are given that the initial position (the distance from home) is 8,000 meters, while the final position is 8,105 meters. We are also given that it’s initially moving at 11 meters per second, as the acceleration during this time frame is 4 meters per squared second. This means that $x=8105$, $x_{0}=8000$, $v_{x}=11$, and $a_{x}=4$. Plug in those values, we will have an equation of $2t^{2}+11t+8000=8105$. Now let’s solve.

First thing’s first. We must subtract $8105$ from both sides to use the zero-product property. By doing this, we should have the equation $2t^{2}+11t-105=0$. When solving these kinds of equations, it’s ideal to use the quadratic formula, but we’re not doing that this time. Instead, we’re using factoring.

Now what two factors of $105$ can you multiply into $1$ and $2$ to generate a difference of $11$? To give out the factors for $105$, they are:

• $1, 3, 5, 7, 15, 21, 35, 105$

Meanwhile, the only two factors of $2$ are $1$ and $2$.

Now let’s find the combination of differences that yield $11$.

• $2\cdot 105-1\cdot 1=209$
• $1\cdot 105-2\cdot 1=103$
• $2\cdot 35-1\cdot 3=67$
• $1\cdot 35-2\cdot 3=29$
• $2\cdot 21-1\cdot 5=37$
• $1\cdot 21-2\cdot 5=11$
• $2\cdot 15-1\cdot 7=23$
• $1\cdot15-2\cdot 7=1$

It seems that $21$ and $5$ are the factors we’re looking for. The polynomial becomes $(2t+21)(t-5)$.

Now let’s solve the equation $(2t+21)(t-5)=0$. Using the zero-product property, either $2t+21=0$ or $t-5=0$. The first equation becomes $t=-\frac{21}{2}$, but since time only moves forward, we’ll disregard that. But $t-5=0$ becomes $t=5$. Our solution is $t=5$.

$2t^{2}+11t-105$
$a=6; b=-23; c=20$

Factors of $2$:
- $1, 2$

Factors of $105$:
- $1, 3, 5, 7, 15, 21, 35, 105$

- $2\cdot 105-1\cdot 1=209$  X
- $1\cdot 105-2\cdot 1=103$  X
- $2\cdot 35-1\cdot 3=67$    X
- $1\cdot 35-2\cdot 3=29$    X
- $2\cdot 21-1\cdot 5=37$    X
- $1\cdot 21-2\cdot 5=11$    ✓
- $2\cdot 15-1\cdot 7=23$    X
- $1\cdot15-2\cdot 7=1$     X

$(2t+21)(t-5)$

1. Use Position Formula, $x=x_{0}+v_{x}t+\frac{1}{2}a_{x}t^{2}$
2. Initial position is $8000$, final position
   is $8105$, initial velocity is $11$,
   acceleration is $4$.
3. $2t^{2}+11t+8000=8105$
4. $2t^{2}+11t-105=0$
5. $(2t+21)(t-5)=0$
6. $2t+11=0, t-5=0$
7. $t=-\frac{21}{2}, t=5$

It takes five seconds for the car to go from 8,000 meters to 8,105 meters.

Quiz #17:

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 “$a=-\mathrm{3,3}$“. Please leave all answers in decimal notation to two digits. So if $a=\frac{5}{3}$, you may use $a=1.67$ or $1.67$, but not $1.7$, $1.667$, or $5/3$.