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

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?

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?

5x^55

Now we have

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

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

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!