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Elza [17]
3 years ago
14

Can someone please help with this or explain how to do it. I'm not very sure where to begin ToT

Mathematics
1 answer:
larisa86 [58]3 years ago
5 0

Answer:

d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

Step-by-step explanation:

First, we find the greatest common factor of 12x^23y^40z^9 and 10x^78z^5.

We work on the numbers, then on x, then on y, then on z.

We start with the GCF of 12 and 10.

10 = 2 * 5

12 = 2 * 3 * 3

The only factor 10 and 12 have in common is 2, so the GCF of 12 and 10 is 2.

Now we work on x. We have x^23 and x^78.

x^23 has 23 factors of x.

x^78 has 78 factors of x.

Since x^23 and x^78 have both at least 23 factors of x, x^23 is the GCF of x.

Now we work on y.

There is y^40 in one term, but there is no y in the other term. There is no GCF for the y.

Now we work on z. The first term has z^9. the second term has z^5. They both have at least 5 factors of z, so the GCF of z is z^5.

Now we put all the parts of the GCF together, and the GCF of the two terms is

2x^23z^5

We use option d. to factor.

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(___a____ + ____b___)

We already have the GCF outside the parentheses. Now we need to figure out what goes inside the parentheses.

The terms that go in positions a and b are the terms that when multiplied by the GCF give you the original expression.

We now work on position a.

What do you multiply by 2x^23z^5 to get 12x^23y^40z^9?

Answer: 6y^40z^4

Now we have

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + ____b___)

Now we do the same for position b.

What do you multiply by 2x^23z^5 to get 10x^78z^5?

Answer: 5x^55

Now we have

12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

Answer: d. 12x^23y^40z^9 + 10x^78z^5 = 2x^23z^5(6y^40z^4 + 5x^55)

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