Factoring a quadratic equation means expressing it as the product of its linear factors.
The standard form of a quadratic equation is ax2+bx+c=0a x squared plus b x plus c equals 0
𝑎𝑥2+𝑏𝑥+𝑐=0
, where aa
𝑎
, bb
𝑏
, and cc
𝑐
are coefficients. The following methods can be used to find its roots (the values of xx
𝑥
that satisfy the equation).
1. Factoring by sum and product (for a=1a equals 1
𝑎=1
)
This method is the most straightforward for simple quadratics where the leading coefficient, aa
𝑎
, is 1.
Steps
-
**Identify bb
𝑏
and cc
𝑐** : Given the equation x2+bx+c=0x squared plus b x plus c equals 0
𝑥2+𝑏𝑥+𝑐=0
.
-
Find two numbers: Look for two numbers, pp
𝑝
and qq
𝑞
, such that their sum equals bb
𝑏
and their product equals cc
𝑐
.
-
p+q=bp plus q equals b
𝑝+𝑞=𝑏
-
p⋅q=cp center dot q equals c
𝑝⋅𝑞=𝑐
-
-
Rewrite the equation: The factored form of the equation is then (x+p)(x+q)=0open paren x plus p close paren open paren x plus q close paren equals 0
(𝑥+𝑝)(𝑥+𝑞)=0
.
-
**Solve for xx
𝑥** : Use the Zero Product Property, which states that if a product is zero, at least one of the factors must be zero.
-
x+p=0⇒x=−px plus p equals 0 implies x equals negative p
𝑥+𝑝=0⇒𝑥=−𝑝
-
x+q=0⇒x=−qx plus q equals 0 implies x equals negative q
𝑥+𝑞=0⇒𝑥=−𝑞
-
Example: Factor x2+7x+10=0x squared plus 7 x plus 10 equals 0
𝑥2+7𝑥+10=0
-
**Identify bb
𝑏
and cc
𝑐** : b=7b equals 7
𝑏=7
, c=10c equals 10
𝑐=10
.
-
Find two numbers: We need two numbers that multiply to 10 and add to 7. The numbers are 5 and 2.
-
5⋅2=105 center dot 2 equals 10
5⋅2=10
-
5+2=75 plus 2 equals 7
5+2=7
-
-
Rewrite the equation: The factored form is (x+5)(x+2)=0open paren x plus 5 close paren open paren x plus 2 close paren equals 0
(𝑥+5)(𝑥+2)=0
.
-
**Solve for xx
𝑥** :
-
x+5=0⇒x=-5x plus 5 equals 0 implies x equals negative 5
𝑥+5=0⇒𝑥=−5
-
x+2=0⇒x=-2x plus 2 equals 0 implies x equals negative 2
𝑥+2=0⇒𝑥=−2
-
2. Factoring by grouping (the AC method)
This method is useful for factoring quadratics where the leading coefficient, aa
𝑎
, is not 1.
Steps
-
Find the "master product": Multiply the coefficients aa
𝑎
and cc
𝑐
to get the product aca c
𝑎𝑐
.
-
Find two numbers: Find two numbers, pp
𝑝
and qq
𝑞
, that multiply to aca c
𝑎𝑐
and add to bb
𝑏
.
-
Split the middle term: Rewrite the original equation by splitting the bxb x
𝑏𝑥
term into pxp x
𝑝𝑥
and qxq x
𝑞𝑥
.
-
ax2+bx+c=ax2+px+qx+ca x squared plus b x plus c equals a x squared plus p x plus q x plus c
𝑎𝑥2+𝑏𝑥+𝑐=𝑎𝑥2+𝑝𝑥+𝑞𝑥+𝑐
-
-
Factor by grouping: Group the first two terms and the last two terms, and factor out the greatest common factor (GCF) from each pair.
-
Factor out the binomial: The two grouped terms should now share a common binomial factor. Factor this binomial out to get the final factored form.
-
**Solve for xx
𝑥** : Use the Zero Product Property to find the roots.
Example: Factor 2x2+9x+10=02 x squared plus 9 x plus 10 equals 0
2𝑥2+9𝑥+10=0
-
Master product: a⋅c=2⋅10=20a center dot c equals 2 center dot 10 equals 20
𝑎⋅𝑐=2⋅10=20
.
-
Find two numbers: Find two numbers that multiply to 20 and add to 9. The numbers are 4 and 5.
-
Split the middle term: 2x2+4x+5x+10=02 x squared plus 4 x plus 5 x plus 10 equals 0
2𝑥2+4𝑥+5𝑥+10=0
.
-
Factor by grouping: (2x2+4x)+(5x+10)=0⇒2x(x+2)+5(x+2)=0open paren 2 x squared plus 4 x close paren plus open paren 5 x plus 10 close paren equals 0 implies 2 x open paren x plus 2 close paren plus 5 open paren x plus 2 close paren equals 0
(2𝑥2+4𝑥)+(5𝑥+10)=0⇒2𝑥(𝑥+2)+5(𝑥+2)=0
.
-
Factor out the binomial: (2x+5)(x+2)=0open paren 2 x plus 5 close paren open paren x plus 2 close paren equals 0
(2𝑥+5)(𝑥+2)=0
.
-
**Solve for xx
𝑥** :
-
2x+5=0⇒x=-5/22 x plus 5 equals 0 implies x equals negative 5 / 2
2𝑥+5=0⇒𝑥=−5/2
-
x+2=0⇒x=-2x plus 2 equals 0 implies x equals negative 2
𝑥+2=0⇒𝑥=−2
-
3. Completing the square
This method transforms a quadratic equation into a perfect square trinomial, allowing it to be solved by taking the square root of both sides.
Steps
-
Move the constant: Rearrange the equation so that the constant term (cc
𝑐
) is on the right side.
-
**Divide by aa
𝑎** : If a≠1a is not equal to 1
𝑎≠1
, divide all terms by aa
𝑎
.
-
Add to both sides: Take the coefficient of the xx
𝑥
term (b/ab / a
𝑏/𝑎
), divide it by 2, and square the result. Add this value to both sides of the equation.
-
Factor the perfect square: The left side of the equation is now a perfect square trinomial and can be factored as a squared binomial, (x+p)2open paren x plus p close paren squared
(𝑥+𝑝)2
.
-
Solve with square roots: Take the square root of both sides, remembering to account for both positive and negative roots (±plus or minus
±
).
-
**Isolate xx
𝑥** : Solve for xx
𝑥
.
Example: Factor x2+6x−7=0x squared plus 6 x minus 7 equals 0
𝑥2+6𝑥−7=0
-
Move the constant: x2+6x=7x squared plus 6 x equals 7
𝑥2+6𝑥=7
.
-
**Divide by aa
𝑎** : The leading coefficient is already 1.
-
Add to both sides: Half of 6 is 3, and 32=93 squared equals 9
32=9
. Add 9 to both sides: x2+6x+9=7+9x squared plus 6 x plus 9 equals 7 plus 9
𝑥2+6𝑥+9=7+9
.
-
Factor the perfect square: (x+3)2=16open paren x plus 3 close paren squared equals 16
(𝑥+3)2=16
.
-
Solve with square roots: x+3=±16⇒x+3=±4x plus 3 equals plus or minus the square root of 16 end-root implies x plus 3 equals plus or minus 4
𝑥+3=±16√⇒𝑥+3=±4
.
-
**Isolate xx
𝑥** :
-
x+3=4⇒x=1x plus 3 equals 4 implies x equals 1
𝑥+3=4⇒𝑥=1
-
x+3=-4⇒x=-7x plus 3 equals negative 4 implies x equals negative 7
𝑥+3=−4⇒𝑥=−7
-
4. The quadratic formula
The quadratic formula is a universal method that can solve any quadratic equation, regardless of whether it is easily factorable. The formula is derived from the method of completing the square.
The formulaFor a quadratic equation in the form ax2+bx+c=0a x squared plus b x plus c equals 0
𝑎𝑥2+𝑏𝑥+𝑐=0
, the roots are given by:x=−b±b2−4ac2ax equals the fraction with numerator negative b plus or minus the square root of b squared minus 4 a c end-root and denominator 2 a end-fraction
𝑥=−𝑏±𝑏2−4𝑎𝑐√2𝑎
Steps
-
**Identify a,b,ca comma b comma c
𝑎,𝑏,𝑐** : Write the equation in standard form, ax2+bx+c=0a x squared plus b x plus c equals 0
𝑎𝑥2+𝑏𝑥+𝑐=0
, and identify the values of the coefficients.
-
Substitute into the formula: Plug the values of aa
𝑎
, bb
𝑏
, and cc
𝑐
into the quadratic formula.
-
Simplify and solve: Calculate the two possible values for xx
𝑥
.
Example: Factor 3x2−x−14=03 x squared minus x minus 14 equals 0
3𝑥2−𝑥−14=0
-
**Identify a,b,ca comma b comma c
𝑎,𝑏,𝑐** : a=3a equals 3
𝑎=3
, b=-1b equals negative 1
𝑏=−1
, c=-14c equals negative 14
𝑐=−14
.
-
Substitute:x=−(-1)±(-1)2−4(3)(-14)2(3)x equals the fraction with numerator negative open paren negative 1 close paren plus or minus the square root of open paren negative 1 close paren squared minus 4 open paren 3 close paren open paren negative 14 close paren end-root and denominator 2 open paren 3 close paren end-fraction
𝑥=−(−1)±(−1)2−4(3)(−14)√2(3)
x=1±1−(-168)6x equals the fraction with numerator 1 plus or minus the square root of 1 minus open paren negative 168 close paren end-root and denominator 6 end-fraction
𝑥=1±1−(−168)√6
x=1±1696x equals the fraction with numerator 1 plus or minus the square root of 169 end-root and denominator 6 end-fraction
𝑥=1±169√6
x=1±136x equals the fraction with numerator 1 plus or minus 13 and denominator 6 end-fraction
𝑥=1±136
-
Simplify and solve:
-
x=1+136=146=73x equals the fraction with numerator 1 plus 13 and denominator 6 end-fraction equals fourteen-sixths equals seven-thirds
𝑥=1+136=146=73
-
x=1−136=-126=-2x equals the fraction with numerator 1 minus 13 and denominator 6 end-fraction equals negative 12 over 6 end-fraction equals negative 2
𝑥=1−136=−126=−2
-
5. Special product patterns
Recognizing special quadratic forms can allow for quick factorization using algebraic identities.
Difference of squares
When a quadratic is a difference of two squares, it takes the form a2x2−b2a squared x squared minus b squared
𝑎2𝑥2−𝑏2
.
-
Factored form: (ax−b)(ax+b)open paren a x minus b close paren open paren a x plus b close paren
(𝑎𝑥−𝑏)(𝑎𝑥+𝑏)
.
-
Example: 9x2−16=09 x squared minus 16 equals 0
9𝑥2−16=0
-
This is a difference of squares: (3x)2−42=0open paren 3 x close paren squared minus 4 squared equals 0
(3𝑥)2−42=0
.
-
Factored form: (3x−4)(3x+4)=0open paren 3 x minus 4 close paren open paren 3 x plus 4 close paren equals 0
(3𝑥−4)(3𝑥+4)=0
.
-
Roots: x=4/3x equals 4 / 3
𝑥=4/3
and x=-4/3x equals negative 4 / 3
𝑥=−4/3
.
-
Perfect square trinomials
These are quadratics that are the result of squaring a binomial.
-
Form: a2x2+2abx+b2a squared x squared plus 2 a b x plus b squared
𝑎2𝑥2+2𝑎𝑏𝑥+𝑏2
or a2x2−2abx+b2a squared x squared minus 2 a b x plus b squared
𝑎2𝑥2−2𝑎𝑏𝑥+𝑏2
.
-
Factored form: (ax+b)2open paren a x plus b close paren squared
(𝑎𝑥+𝑏)2
or (ax−b)2open paren a x minus b close paren squared
(𝑎𝑥−𝑏)2
.
-
Example: x2−10x+25=0x squared minus 10 x plus 25 equals 0
𝑥2−10𝑥+25=0
-
This is a perfect square trinomial: (x)2−2(x)(5)+52=0open paren x close paren squared minus 2 open paren x close paren open paren 5 close paren plus 5 squared equals 0
(𝑥)2−2(𝑥)(5)+52=0
.
-
Factored form: (x−5)2=0open paren x minus 5 close paren squared equals 0
(𝑥−5)2=0
.
-
Root: x=5x equals 5
𝑥=5
.
-
Choosing the right method
-
Sum and Product: The fastest method for simple quadratics where a=1a equals 1
𝑎=1
.
-
Grouping: Best for quadratics where a≠1a is not equal to 1
𝑎≠1
but the numbers are relatively easy to find.
-
Completing the Square: A good technique for understanding the structure of quadratics and for deriving the quadratic formula, but often more work than the other methods.
-
Quadratic Formula: The most reliable method, as it works for all quadratic equations, including those with irrational or complex roots.
-
Special Forms: Use these for quick factorization when the equation matches a known pattern.