Find Polynomial Function...2

Section 2.5
Question 48



We are given 3 zeros. This leads to a 4th degree equation. Why fourth degree?

If x = -1 is a zero, then (x - 1) is a factor.

If x = 2 is zero, then (x + 2) is a zero.

If x = sqrt{2}I is a zero, then (x + 2sqrt{2}i) is a factor.

f(x) = a(x - 1)(x + 2)(x - sqrt{2}i)(x + sqrt{2}i)

After further simplification, I got the following:

f(x) = a(x^2 + x - 2)(x^2 + 2)

We are told that f(0) = 4.

To find a, let x = 0 and set equation to 4. Solve for a.

Doing this, I got the following:

a = -1.

f(x) = -(x^2 + x - 2)(x^2 + 2)

f(x) = -x^4 - x^3 - 2x + 4

You say?

 

you made mistake with factors

We are given:
4th degree equation
zeros: -1, 2, sqrt(2)*i -> complex roots always come in pairs, so you also have fourth zero which is -sqrt(2)*i
that is why is given “fourth degree “

if zero -1, factor is (x-(-1))=(x+1)
if zero 2, factor is (x-2)
if zero sqrt(2)i, factor is (x - sqrt(2)i)
if zero -sqrt(2)i, factor is (x + sqrt(2)i)

given also f(1) =12 => when x=1 then f(x)=12

f(x) = a(x +1)(x - 2)(x - sqrt(2)i)(x + sqrt(2)i)

f(x) = a(x^4 - x^3 - 2x - 4) ..........now substitute x=1 and f(x)=12 to calculate a

12= a(1^4 - 1^3 - 2 *1 - 4)

12= a(-6)
a=12/-6
a=-2

then

f(x) = -2(x^4 - x^3 - 2x - 4)

f(x) =-2x^4 + 2x^3 + 4x + 8

check solution point:

f(1) = -2×1^4 + 2×1^3 + 4×1 + 8=12

 

What did I do wrong here? Can you check a thread posted yesterday at 9:16pm entitled Find Zeros of Polynomial Functions?

 

when you came here f(x) = a(x^2 + x - 2)(x^2 + 2)
you say: We are told that f(0) = 4. => this was wrong

you needed to use solution point f(1) =12

 

I used the wrong solution point. Thanks.