Factoring binomials of the form x2−y2=(x−y)(x+y):
This approach applies the difference of two squares identity.
Factor x2−16.
We have
x2−16=x2−42=(x−4)(x+4). □
Factor x4−25y4.
We have
x4−25y4=(x2)2−(5y2)2=(x2−5y2)(x2+5y2). □
Factor 9x3y−36xy.
We have
9x3y−36xy=9xy(x2−4)=9xy(x2−22)=9xy(x−2)(x+2). □
Factor (x+1)2−9(x−2)2.
We have
(x+1)2−9(x−2)2=(x+1)2−(3(x−2))2=((x+1)−3(x−2))((x+1)+3(x−2))=(x+1−3x+6)(x+1+3x−6)=(−2x+7)(4x−5). □
Factoring binomials of the form x3+y3=(x+y)(x2−xy+y2) and x3−y3=(x−y)(x2+xy+y2):
This approach applies the sum and difference of cubes identity.
Factor 27x3−y3.
We have
27x3−y3=(3x)3−y3=(3x−y)((3x)2+3xy+y2)=(3x−y)(9x2+3xy+y2). □
Factor 64x3+27y3.
We have
64x3+27y3=(4x)3+(3y)3=(4x+3y)((4x)2−12xy+(3y)2)=(4x+3y)(16x2−12xy+9y2). □
2.32+0.69+0.092.33−0.027
Quick! Calculate the expression above as fast as possible! Time is of the essence!
Factoring trinomials of the form x2+2xy+y2=(x+y)2 and x2−2xy+y2=(x−y)2:
Factor x2+6x+9.
We have
x2+6x+3=x2+2(x⋅3)+32=(x+3)2. □
Factor x2−4x+4.
We have
x2−4x+4=x2−2(x⋅2)+22=(x−2)2. □
Factor 25x2−20x+4.
We have
25x2−20x+4=(5x)2−2(5x⋅2)+22=(5x−2)2. □
Factor 16x2−24xy+9y2.
We have
16x2−24xy+9y2=(4x)2−2(4x⋅3y)+(3y)2=(4x−3y)2. □
Factoring trinomials of the form x2+(a+b)x+ab=(x+a)(x+b):
This approach is also known as factorization by observation.
In cases where we don't have a perfect square of the form (x+y)2 or (x−y)2, but the leading coefficient of x is 1, we can try to find a and b such that a⋅b is equal to the constant term, and a+b is equal to the coefficient of x.
Factor x2+5x+6.
We have
x2+5x+6=x2+(2+3)x+(2×3)=(x+2)(x+3). □
Factor x2−3x−4.
We have
x2−3x+4=x2+(1−4)x+(1×(−4))=(x+1)(x−4). □
Factor x2−12x+35.
We have
x2−12x+35=x2+(−5−7)x+((−5)×(−7))=(x−5)(x−7). □
Factor 2x2+4xy−30y2.
We have
2x2+4xy−30y2=2(x2+2xy−15y2)=2(x2+(5y−3y)x+(5y)×(−3y))=2(x+5y)(x−3y). □
Factoring trinomials of the form acx2+(ad+bc)x+bd=(ax+b)(cx+d):
When the leading coefficient of x is not 1, we must factor both the leading coefficient and the constant.
Factor 2x2+4x+2.
Factors of 2x2: 2x and x
Factors of 2: 1 and 2
Since we're looking at all-positive coefficients, we can ignore negative factors.
Given this, two simple possible factorizations are (2x+1)(x+2) and (2x+2)(x+1) .
In both of these factorizations, the leading term will be 2x⋅x=2x2 and the constant term will be 1⋅2=2, So we don't have to worry about them. Let's turn our attention to getting the correct middle term 4x.
In the factorization (2x+1)(x+2), the middle term would be (2x⋅2)+(1⋅x)=5x, which is incorrect.
In the factorization (2x+2)(x+1), the middle term would be (2x⋅1)+(2⋅x)=4x, which is what we want. So the factorization that gives us the correct expansion is (2x+2)(x+1). □
Factor 2xy5−11xy4−6xy3.
We can factor out xy3 to make our work simpler:
2xy5−11xy4−6xy3=xy3(2y2−11y−6).
Now we can just concentrate on factoring 2y2−11y−6.
Factors of 2y2: 2y and y, −2y and −y
Factors of −6: −2 and 3, 2 and −3, −1 and 6, 1 and −6
There are many different combinations we can try for the factorization. But one thing to notice is that since the middle term is −11y, we need to use factors that can get us to the number −11: 2y×(−6) and y×1 would give us the middle term we want.
Since we want 2y to be multiplied by −6 to give us the middle term, the −6 must exist in a different set of parentheses from 2y so that it can be distributed to and multiplied by 2y.
So let's check the factorization (2y+1)(y−6). The leading term is 2y⋅y=2y2, and the constant term is 1⋅−6=−6 as desired. The middle term is (2y⋅(−6))+1⋅y=−11y, as we wanted.
So 2y2−11y−6 = (2y+1)(y−6), implying that our final answer is
2xy5−11xy4−6xy3=xy3(2y2−11y−6)=xy3(2y+1)(y−6). □
Factoring polynomials in this way involves some amount of guessing and checking. You can greatly improve your speed at this process by using your number sense to figure out which combinations of numbers will successfully get you the middle term that you want.