Factoring Quadratic Equations
Factoring is a method for solving quadratic equations.
This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Factoring Quadratic Equations. In Justin Math: Algebra. https://justinmath.com/factoring-quadratic-equations/
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Factoring is a method for solving quadratic equations. It involves converting the quadratic equation to standard form, then factoring it into a product of two linear terms (called factors), and finally solving for the variable values that make either factor equal to 0.
When we factor, we are rearranging the equation to say that the product of two numbers is 0. The equation is solved when either number is 0, because any number multiplied by 0 is 0.
How to Factor
Factoring is easiest in hindsight. Multiplying through, we see that the factored form is equivalent to the standard form:
But how can we know this to begin with? In other words, if we want to factor an expression x2+bx+c into the form (x+m)(x+n), how do we know what m and n are?
Hereβs the trick: m and n need to multiply to c and add to b.
To factor the expression x2+5x+4, we need to find two numbers that multiply to 4 and add to 5. Although 2 and 2 multiply to 5, they donβt add to β2. But 1 and β3 multiply to β3 AND add to β2, so they work! The factored form is then (x+1)(xβ3).
Even with negatives, the method is still the same: to factor the expression x2β2xβ3, we need to find two numbers that multiply to β3 and add to β2. Although β1 and 3 multiply to β3, they donβt add to β2. But 1 and β3 multiply to β3 AND add to β2, so they work! The factored form is then (x+1)(xβ3).
Case of Many Potential Factors
Factoring can become a little tricky when c has a lot of factors. In such cases, it can be helpful to make a factor table.
For example, to factor x2+26x+144, we can list out the factors of 144 and find which pair adds to 26. Since this pair is 8 and 18, the expression factors to (x+8)(x+18).
To speed up the process, notice that the sums are automatically ordered from biggest to smallest β so we donβt necessarily have to create the whole table.
We could have started with some intermediate pair, say 6 and 24, and realized that since the sum is too big, we need the first factor to be bigger than 6.
Or, we could have noticed that sum of 12 and 12 is in the ballpark of 26, and worked our way up from the bottom of the table.
Case of Negative Terms
To deal with a negative value for b, we could use the same method as before, except that we would have to make both factors negative.
For example, since we know that 8 and 18 are factors of 144 that add to 26, we also know that β8 and β18 are factors of 144 that add to β26, so the expression x2β26x+144 factors to (xβ8)(xβ18).
To deal with a negative value for c, we can think about the difference instead of the sum.
For example, to factor x2+32xβ144, we can find which factor pair of 144 has a difference of 32, and put a negative on the smaller factor to make the sum. Since this pair is 4 and 36, the expression factors to (xβ4)(x+36).
If b were negative as well β say, if we wanted to factor x2β32xβ144 β then we could use the same process but put the negative on the bigger factor to make the sum negative. That is, we would put the negative on the 36 instead of the 4, and the resulting factored form would then be (x+4)(xβ36).
Case of a Common Factor
Sometimes, we can simplify quadratic expressions by factoring out something that ALL the terms have in common.
This makes it easy to factor quadratic expressions where c is 0 β just factor out the variable!
Case when the Leading Coefficient is Not One
Factoring out the variable works even when a is something other than 1.
But what about when a is something other than 1, and c is not zero?
Thereβs a little trick that lets us reduce this to a factoring problem with a equal to 1. We multiply c by a, replace a with 1, factor the result, divide each constant in each factor by the original a, and move denominators onto our variables.
Weβll talk about why this trick works in the next chapter, when we cover the quadratic formula.
Case of No Middle Term
Lastly, what about when b is 0? Since the factors have to add to b, they must be negatives of each other. Since the factors have to multiply to c, and they are the same number (except one is negative), they must be the positive and negative square roots of c!
For example, x2β4 factors to (x+2)(xβ2), and x2β9 factors to (x+3)(xβ3).
This trick also works if a is not equal to 1 β we just have to factor a out first.
Exercises
Factor the following quadratic equations. Then, use the factored form to find the solutions. (You can view the solution by clicking on the problem.)
1)x2+7x+12=0
Solution:
(x+3)(x+4)=0
x=β3 or x=β4
2)x2+9x+14=0
Solution:
(x+7)(x+2)=0
x=β7 or x=β2
3)x2β7x=β10
Solution:
(xβ2)(xβ5)=0
x=2 or x=5
4)x2+18=9x
Solution:
(xβ6)(xβ3)=0
x=6 or x=3
5)x2+2x=8
Solution:
(x+4)(xβ2)=0
x=β4 or x=2
6)21β4x=x2
Solution:
(x+7)(xβ3)=0
x=β7 or x=3
7)3x+10=x2
Solution:
(x+2)(xβ5)=0
x=β2 or x=5
8)x2β5x=36
Solution:
(x+4)(xβ9)=0
x=β3 or x=9
9)4x2=β52x
Solution:
4x(x+13)=0
x=0 or x=β13
10)β8x2+64x=0
Solution:
8x(xβ8)=0
x=0 or x=8
11)x2β25=0
Solution:
(x+5)(xβ5)=0
x=β5 or x=5
12)x2β144
Solution:
(x+12)(xβ12)=0
x=β12 or x=12
13)12x2+11x=β2
Solution:
(4x+1)(3x+2)=0
x=β14 or x=β23
14)10x2=27xβ5
Solution:
(2xβ5)(5xβ1)=0
x=52 or x=15
15)5x=4β6x2
Solution:
(3x+4)(2xβ1)=0
x=β43 or x=12
16)21x2β10=β29x
Solution:
(7xβ2)(3x+5)=0
x=27 or x=β53
This post is part of the book Justin Math: Algebra. Suggested citation: Skycak, J. (2018). Factoring Quadratic Equations. In Justin Math: Algebra. https://justinmath.com/factoring-quadratic-equations/
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