SOLUTION: The polynomial P(x)=x^4+ax^2+bx-9 has zeros at x=1 and x=-3. The values of a and b are...

Question 968069: The polynomial P(x)=x^4+ax^2+bx-9 has zeros at x=1 and x=-3. The values of a and b are...
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
if the polynomial has zeroes at x = 1 and x = -3, then the value of the polynomial would have to be 0 when you replace x with either 1 or -3.

you would get:

f(1) = 0 and f(-3) = 0

start with your polynomial and set it to 0.

you will get:

x^4+ax^2+bx-9 = 0

replace x with 1 and the equation becomes:

1^4 + a*1 + b*1 - 9 = 0
simplify to get:
1 + a + b - 9 = 0
combine like terms to get:
a + b - 8 = 0
add 8 to both sides of the equation to get:
a + b = 8

replace x with -3 and the equation becomes:
(-3)^4 + a*(-3)^2 + b*(-3) - 9 = 0
simplify this to get:
81 + 9a - 3b - 9 = 0
combine like terms to get:
72 + 9a - 3b = 0
subtract 72 from both sides of the equation to get:
9a - 3b = -72

you have 2 equations that need to be solved simultaneously.
they are:
a + b = 8
9a - 3b = -72

multiply both sides of the first equation by 9 to get:

9a + 9b = 72
9a - 3b = -72

subtract the second equation from the first to get:

12b = 144

solve for b to get b = 12

go back to the first equation and solve for a to get:
a + b = 8 becomes:
a + 12 = 8
solve for a to get:
a = 8 - 12 = -4

you have:

a = -4
b = 12

replace those values in the original equation of x^4+ax^2+bx-9 = 0 to get:
x^4 -4 * x^2 + 12 * x - 9 = 0

if you graph the equation using a graphing calculator that tells you the zero points, your graph will look like this:

$$$

another way would have been to evaluate the equation at f(1) and f(-3) when the value of a was -4 and the value of b was 12.

f(1) and f(-3) should both have been equal to 0.

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