SOLUTION: I don't understand how to get the final answer to this type of problem. Fourth degree, zeros of -5,-2,1, and 3, and y-intercept of 15. p(x)=a(x+5)(x+2)(x-1)(x-3) p(0)=a(5)(2

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Question 167067This question is from textbook College Algebra
: I don't understand how to get the final answer to this type of problem.
Fourth degree, zeros of -5,-2,1, and 3, and y-intercept of 15.
p(x)=a(x+5)(x+2)(x-1)(x-3)
p(0)=a(5)(2)(-1)(-3)=30a
30a=15 a=1/2
p(x)=1/2(x+5)(x+2)(x-1)(x-3) How do you mutiply out the four linear factors and the constant?
This question is from textbook College Algebra

Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Distributive property
Let's look at Z(C+D)=ZC+ZD.
What if Z=A+B?
Then
Z(C+D)=(A+B)(C+D)=ZC+ZD=(A+B)C+(A+B)D=AC+BC+AD+BD
So then,

and

This is the FOIL method for expanding quadratic equations from factors.
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.
.
Now the problem becomes,

Let's look at the product of the two quadratics,
.
.
.
We again have to use the distributive property and make sure we account for each term.
There are nine terms in all.





Now we'll put that back in our original function.


.
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Check, as an exercise, that we didn't make any mistakes and that x= -5,-2,1, and 3 are in fact zeros of this polynomial.

So that checks out.

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