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

Algebra ->  Polynomials-and-rational-expressions -> 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      Log On


   

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) About Me  (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,
%28x%2B5%29%28x%2B2%29=x%2Ax%2B5x%2B2x%2B10=x%5E2%2B7x%2B10
and
%28x-1%29%28x-3%29=x%2Ax%2B%28-3%29x%2B%28-1%29x%2B3=x%5E2-4x%2B3
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,
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We again have to use the distributive property and make sure we account for each term.
There are nine terms in all.




%28x%5E2%2B7x%2B10%29%28x%5E2-4x%2B10%29=%0D%0Ax%5E4%2B3x%5E3-15x%5E2-19x%2B30
Now we'll put that back in our original function.
p%28x%29=%281%2F2%29%28x%5E4%2B3x%5E3-15x%5E2-19x%2B30%29
p%28x%29=%281%2F2%29x%5E4%2B%283%2F2%29x%5E3-%2815%2F2%29x%5E2-%2819%2F2%29x%2B15%29
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.
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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.
p%280%29=15
So that checks out.