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# I need to factor theses polynomials 9x^2-16 anda^3-64

I need to factor these two polynimials :9x^2-16 anda^3-64

### 4 Answers by Expert Tutors

Jordan K. | Nationally Certified Math Teacher (grades 6 through 12)Nationally Certified Math Teacher (grade...
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The polynomials, 9x2 - 16 and a3 - 64 are the differences of two perfect squares and two perfect cubes, respectively.

Let's begin by recognizing the factoring pattern for the difference of two perfect squares:
a2 - b2 = (a + b)(a - b)
Let's apply this factoring pattern to our polynomial, 9x2 - 64, substuting 9x2 for a2 and 16 for b2 which gives us 3x for a and 4 for b.
Plugging in these values for a and b, our factored expression is (3x + 4)(3x - 4).

Now let's consider the factoring pattern for the difference of two perfect cubes:
a3 - b3 = (a - b)(a2 + ab + b2)
Let's use this factoring pattern to factor our polynomial, a3 - 64, substituting a for a and 4 for b.
Plugging in thesen values for a and b, our factored expression is (a - 4)(a2 + 4a + 16).

Roman C. | Masters of Education Graduate with Mathematics ExpertiseMasters of Education Graduate with Mathe...
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Most generally the difference of two n-th powers factors as,

an - bn = (a - b)(an-1 + an-2b+ an-3b2... + a2bn-3 + abn-2 + bn-1)

If n is composite, then the second factor can be factored further, which is a straight forward exercise using the distributive property.

9x2 - 16 = (3x)2 - 42= (3x - 4)(3x + 4)

a3 - 64 = a3 - 43 = (a - 4)(a2 + 4a + 42) = (a - 4)(a2 + 4a + 16)

Shefali J. | A Complete Math Tuition SolutionA Complete Math Tuition Solution
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Hello Tanesha,

First we'll factorize 9x2 - 16. If you look at the problem it is based on factoring a difference of two squares

[(a2 - b2) = (a + b) (a - b)]. Note that the sum of two squares DOES NOT factor.

When you have the difference of two bases being squared, it factors as the product of the sum and the difference of the bases that are being squared.

9x2 - 16

= (3x)2 - (4)2(form of a difference of two squares)

= (3x + 4)(3x - 4) (factor as the product of the sum and the differnce of the bases)(answer)

To check your answer multiply (3x + 4)(3x -4) using FOIL method and you'll get back 9x2 - 16.

Now let's do a3 - 64. As you can see that it is factoring a difference of two cubes

(a3 - b3) = (a - b)(a2 + ab + b2).

When you have the difference of two cubes, you get a product of a binomial and a trinomial. The binomial is the differnce of that bases that are cubed. Trinomial is the first base square, the second term is the opposite of the product of the two bases, and the third term is the second base squared.

a3 - 64

= (a)3 - (4)3 (form of a difference of two cubes)

= (a - 4) [(a)2 + (a)(4) + (4)2] (binomial is difference of bases)

(trinomial is first base squared, plus product of bases, plus second base squared)

= (a - 4)(a2 + 4a + 16) (answer)

To check your answer multiply binomial and trinomial and you'll get back a3 - 64.

Good luck factorizing!

Kathryn D. | Worry-Free, Stress-Free, Katie D. You can count on me.Worry-Free, Stress-Free, Katie D. You ca...
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General Information:

For x± # x ± # you have a ( x ± # )( x ± # ) The two numbers at the end have to add together to get to # x  and have to multiply together to get to the ending #

When you have a number in front of the x2 that number must be split up so that when you multiply ( x ± # )( x ± # ) your "x"s will get you back to that number. So if you have a problem that starts with 6x2 you can have either ( 1x ± # )( 6x ± # ) OR ( 2x ± # )( 3x ± # ) because 1*6=6 and 2*3=6.