I need to factor these two polynimials :9x^2-16 anda^3-64
I need to factor theses polynomials 9x^2-16 anda^3-64
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).
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.
For your cases:
9x2 - 16 = (3x)2 - 42= (3x - 4)(3x + 4)
a3 - 64 = a3 - 43 = (a - 4)(a2 + 4a + 42) = (a - 4)(a2 + 4a + 16)
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!
For x2 ± # 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.
Your first problem:
9x2 - 16
The first thing that we have to worry about it the 9. You can have either (9x ± #)(1x ± #) OR (3x ± #)(3x ± #). What I would usually do from here is figure out what multiplies together to get sixteen: 1*16, 2*8, 4*4, 8*2, and 16*1 (order matters!). Then (becasue we have to take into account the numbers in front of x), I would plug them into each of the equations until I found one where the singular "x"s canceled out. That would give us our 9x2 +0x -16
Because I've been at this a while, I know that 16 is a perfect square (4*4=16), So, I would choose the second of the two potential answer bases (3x ± #)(3x ± #) to use. Because 4 is the same either way, and I know that they need to cancel out in the middle, one would be positive and the other would be negative to look like this: (3x + 4)(3x - 4) then, like any good problem, you should multiply it back out and check that yes (3x + 4)(3x - 4) = 9x2-16