Section 1.5: Factoring Sums and Differences of Squares
#1- 42: Completely factor the binomials, remember to factor out the GCF first when applicable
(if a problem is prime say so).
1) x2 – 9
Divide the exponent by 2 and take the square root of the number. Then create two
basically identical parenthesis, except for opposite signs.
Solution: (x + 3)(x – 3)
2) x2 + 9
This is a sum of squares and it is prime.
Solution: Prime
5) y2 – 36
Divide the exponent by 2 and take the square root of the number. Then create two
basically identical parenthesis, except for opposite signs.
Solution: (y + 6)(y – 6)
7) y2 + 36
This is a sum of squares and it is prime.
Solution: Prime
9) 25a2 – 81
Divide the exponent by 2 and take the square root of the numbers. Then create two
basically identical parenthesis, except for opposite signs.
Solution: (5a + 9)(5a – 9)
11) 25a2 + 81
This is a sum of squares and it is prime.
Solution: Prime
13) 49x2 – 36
Divide the exponent by 2 and take the square root of the numbers. Then create two
basically identical parenthesis, except for opposite signs.
Solution: (7x + 6)(7x – 6)
15) 49x2 + 36
This is a sum of squares and it is prime.
Solution: Prime
17) x3 – 64x
First factor out the common factor of x.
= x(x2 – 64)
Now factor the inside of the parenthesis.
Solution: x(x + 8)(x – 8)
19) x3 + 64x
First factor out the common factor of x. This will be the answer, as what is left inside the
parenthesis doesn’t factor.
Solution: x(x2 + 64)
21) 3x2 – 27
First factor out the common factor of 3.
= 3(x2 – 9)
Now factor the inside of the parenthesis.
Solution: 3(x+3)(x – 3)
23) 3x2 + 27
First factor out the common factor of 3. This will be the answer, as what is left inside the
parenthesis doesn’t factor.
Solution: 3(x2 + 9)
25) 9 – 25x2
This could be rewritten with the -25x2 first. If you do this your answer would be –(5x + 3)(5x-3)
I will leave the problem written in the order it is given. So the answer will look a bit backwards.
Take the square root of the 9 to get 3, the square root of 25 gives 5 and divide the exponent by
2.
Solution: (3 + 5x)(3 – 5x)
27) 81 – 16x2
This could be rewritten with the -16x2 first. If you do this your answer would be –(4x + 9)(4x-9)
I will leave the problem written in the order it is given. So the answer will look a bit backwards.
Take the square root of the 81 to get 9, the square root of 16 gives 4 and divide the exponent
by 2.
Solution: (9 – 4x)(9 + 4x)
29) x4 – 9
You need to divide the exponent of the x by 2.
Solution: (x2 + 3)(x2 – 3)
31) 16x4 – 25
Divide the exponent by 2 and square root the numbers.
Solution: (4x2 + 5)(4x2 – 5)
33) 98y2 – 2x4
First factor out the GCF of 2.
= 2(49y2 – x4)
Next divide the exponents by 2 and square root the number.
Solution: 2(7y + x2)(7y – x2)
35) x4 – 16
Trick problem************
this problem is a trick, because one of the two parenthesis factors after factor the first time.
Step 1 answer: = (x2 + 4)(x2 – 4)
While this is correct, it is not completely factored. The x2 + 4 parenthesis is prime, but the x2 – 4
factors more.
Solution: (x2 + 4)(x + 2)(x – 2)
37) 2x4 – 512
First factor out the GCF of 2.
= 2(x4 – 256)
Now divide the exponent by 2 and square root the 256.
= 2(x2 + 16)(x2 – 16)
Again this is tricky. The above line is correct, but not completely factored. I need to factor the
x2 – 16 parenthesis to get the best final answer.
Solution: 2(x2 + 16)(x + 4)(x – 4)
39) y4 – 2401
I needed to use my calculator to find the square root of 2401.
= (y2 + 49)(y2 – 49)
Now factor the second parenthesis.
Solution: (y2 + 49)(y + 7)(y – 7)
41) x4 + 4
This is a sum of squares, so it is prime.
Solution: Prime