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When a function has a zero at x=p, it has a factor (x-p). When a polynomial function with real coefficients has a complex zero, its conjugate is also a zero.

Factored form

Given the two zeros and the one we can infer, we can factor our 3rd-degree polynomial function as ...

f(x) = a(x -(-4))·(x -(4+5i))·(x -(4-5i))

Real factors

Using the factoring of the difference of squares, we can combine the complex factors to make a real factor.

f(x) = a(x +4)((x -4)² -(5i)²) = a(x +4)(x² -8x +16 +25)

Finding the scale factor

The value of this at x=1 is ...

f(1) = a(1 +4)(1 -8 +41) = 170a

We want f(1) = 170, so ...

170 = 170a ⇒ a=1

The factored polynomial function is ...

f(x) = (x +4)(x² -8x +41)

Standard form

Expanding this expression, we have ...

f(x) = x(x² -8x +41) +4(x² -8x +41) = x³ -8x² +41x +4x² -32x +164

f(x) = x³ -4x² +9x +164

Graph

The attached graph verifies the real zero (x=-4) and the value at x=1. It also shows that the factor with complex roots has vertex form (x -4)² +25, exactly as it should be.

edisonlara1212 edisonlara1212 · Answer 2 · 2022-07-29T04:21:29+02:00

Answer:

[tex]f(x) = (x+4)(x^2-8x+41)[/tex]

Step-by-step explanation:

Ok, so there are a couple of things to note here. The first thing is that there is a complex solution

Complex Conjugate Root Theorem:

if [tex]a-bi[/tex] is a solution then [tex]a+bi[/tex] is a solution and vice versa

Fundamental Theorem Of Algebra:

Any polynomial with a degree "n", will have "n" solutions. Those solutions can be real and imaginary numbers

So since we're given the root: [tex]4+5i[/tex], we can use the Complex Conjugate Root Theorem to assert that: [tex]4-5i[/tex] is also a solution.

So now we know 3 solutions/zeroes, and since n=3 (the degree), we can know for a fact that we have all the solutions due to the Fundamental Theorem of Algebra.

So using these roots, we can express the polynomial as it's factors. When you express a polynomial as factors it'll look something like so: [tex]f(x) = a(x-b)(x-c)(x-d)...[/tex] where a, b, and d are zeroes of the polynomial. Also notice the "a" value? This will affect the stretch/compression of the polynomial.

So let's express the polynomial in factored form:

[tex]f(x) = a(x-(-4))(x-(4+5i))(x-(4-5i))[/tex]

Simplify the x-(-4)

[tex]f(x) = a(x+4)(x-(4+5i))(x-(4-5i))[/tex]

Now let's distribute the negative sign to the complex roots

[tex]f(x) = a(x+4)(x-4-5i)(x-4+5i))[/tex]

Now let's rewrite the two factors (x-4-5i) and (x-4+5i) so the (x-4) is grouped together

[tex]f(x) = a(x+4)((x-4)-5i)((x-4)+5i))[/tex]

If you look at the two complex factors, this looks very similar to the difference of squares: [tex](a-b)(a+b) = a^2-b^2[/tex]

In this case a=(x-4) and b=5i. So let's use this identity to rewrite the two factors

[tex]f(x) = a(x+4)((x-4)^2-(5i)^2)[/tex]

Let's expand out the (x-4)^2

[tex]f(x) = a(x+4)(x^2+2(-4)(x)+(-4)^2-(5i)^2)[/tex]

Simplify

[tex]f(x) = a(x+4)(x^2-8x+16-(5i)^2)[/tex]

Now simplify the (5i)^2 = 5^2 * i^2

[tex]f(x) = a(x+4)(x^2-8x+16-(-25))[/tex]

Simplify the subtraction (cancels out to addition)

[tex]f(x) = a(x+4)(x^2-8x+41)[/tex]

So just to check for the value of "a", we can substitute 1 as x, and set the equation equal to 170

[tex]170 = a(1+4)(1^2-8(1)+41)\\170 = a(5)(1-8+41)\\170 = a(5)(34)\\170 = 170a\\a=1[/tex]

In this case it's just 1, so the polynomial can just be expressed as:
[tex]f(x) = (x+4)(x^2-8x+41)[/tex]

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