# How do you factor or solve polynomials? It says factor each polynomial. 1. 64-40ab 11. 5x^3y^2+10x^2y+25x 12. 9ax^3+18bx^2+24cx If you can please be specific. The teacher explained yesterday and I...

How do you factor or solve polynomials?

It says factor each polynomial.

1. 64-40ab

11. 5x^3y^2+10x^2y+25x

12. 9ax^3+18bx^2+24cx

If you can please be specific. The teacher explained yesterday and I didn't pick it up.

Thank you!!!

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### 2 Answers

To factor polynomials, the principle of prime factorization can be used. Prime factorization is listing the prime factors of the term. When we say prime, the factors are just 1 and itself. You can use the tree diagram for better illustration

Let's apply this for the first, 64-40ab. Start with the least prime number, it's 2. Since the first term do not have variables or letters beside it, we just look for the factors of the numbers. So this is how it looks:

1. 64 - 40ab

64 40

/ \ / \

2 32 2 20

/ \ / \

2 16 2 10

/ \ / \

2 8 2 5

/ \

2 4

/ \

2 2

List down all the PRIME factors of the numbers. It will be better if you'll align the similar numbers. Then bring down the numbers that are both present on each number. If you multiply the common numbers, GCF (greatest common factor) is the result.

64 = 2 * 2 * 2 * 2 * 2 * 2

40 = 2 * 2 * 2 * 5

_________________________

GCF: 2 * 2 * 2 = 8

There are only 3 2's on 40 so we can only consider 3 2's on 60. That's why we have 8 as GCF. Now factor 8 on each term. If you move out or factor 8, you also divide the original polynomial by the GCF.

so, `8 (64/8 - (40ab)/8)`

8(8-5ab) -------> is the factored form of 64 - 40ab.

For the second problem, 5^3y^2+10x^2y+25x.

Bring out 5, since it id the GCF of the numerical coexfficients, 5, 10 and 25. Then for the variable part, x is present on all the terms but y is just in the first two. So we just bring out x. In getting the GCF of variables, just consider the variable with the smallest coeeficient. So 5x is the GCF for the whole polynomial.

`5x((5x^3y)/(5x)+(10x^2y)/(5x)+(25x)/(5x))`

That is the factored form of 5^3y^2+10x^2y+25x. Note that when you divide variables, you just subtract the coeeficients. So x^3/x will be x^2.

And lastly, 9ax^3+18bx^2+24cx. The GCF for the numbers is 3 and x is present to all terms.

So `9ax^3 + 18bx^2 + 24cx = 3x(3ax^2 + 6bx + 8 )`

For 60 - 4ab, we identify the gcf. There is no common factor for the variables. Hence, we only look on the constants 64 and 40.

What is the number that both 64 and 40 are divisible with?

Prime factorizing 64 and 40:

64 = 2*2*2*2 = 2^3 *2

40 = 2*2*2*5 = 2^3 * 5.

Hence, the gcf is 2^3 = 8.

So, we factor out 8:

64 - 40ab = **8(8 - 5ab)**.

For 5x^3y^2+10x^2y+25x, we will look on the constants first.

Its 5, 10 and 25. So, the gcf is 5.

Now, we look on the variables, all of the terms has a variable x with them: x^3, x^2, and x. We will choose the one with the least amount of exponent. So, gcf is x.

Therefore, we factor out 5*x = 5x.

5x^3y^2+10x^2y+25x = **5x(x^2y^2 + 2xy + 5)**.

Lastly for 9ax^3+18bx^2+24cx, we look on 9, 18 and 24.

Prime factorizations are:

9 = 3*3

18 = 3*3 *2

24 = 3*2*2*2

So, gcf = 3.

For the variables, all has x but we choose the least amount of exponent. So, gcf is x.

Hence, we factor out 3*x = 3x.

9ax^3+18bx^2+24cx =** 3x(3ax^2 + 6bx+8c)**.