High School

Factor [tex]f(x)=4x^3+15x^2-184x+45[/tex] into linear factors given that -9 is a zero of [tex]f(x)[/tex].

[tex]f(x)=4x^3+15x^2-184x+45=[/tex] [tex]\square[/tex]

(Factor completely.)

Answer :

Certainly! Let's factor the polynomial [tex]\(f(x) = 4x^3 + 15x^2 - 184x + 45\)[/tex] completely, given that [tex]\(-9\)[/tex] is a zero of the polynomial.

### Step 1: Use the given zero to factor the polynomial.
Since [tex]\(-9\)[/tex] is a zero of the polynomial, [tex]\(x + 9\)[/tex] is a factor of [tex]\(f(x)\)[/tex].

### Step 2: Perform polynomial division.
To factor [tex]\(f(x)\)[/tex], divide it by [tex]\(x + 9\)[/tex] to find the other factor, which will be a quadratic polynomial:

- Divide [tex]\(4x^3 + 15x^2 - 184x + 45\)[/tex] by [tex]\(x + 9\)[/tex].
- The division gives us:

[tex]\[
\frac{f(x)}{x+9} = 4x^2 - 21x + 5
\][/tex]

This means:
[tex]\[
f(x) = (x + 9)(4x^2 - 21x + 5)
\][/tex]

### Step 3: Factor the quadratic polynomial.
Next, we factor the quadratic polynomial [tex]\(4x^2 - 21x + 5\)[/tex].

- To factor [tex]\(4x^2 - 21x + 5\)[/tex], look for two numbers that multiply to [tex]\(4 \times 5 = 20\)[/tex] and add to [tex]\(-21\)[/tex].
- These numbers are [tex]\(-1\)[/tex] and [tex]\(-20\)[/tex].
- Rewrite the quadratic as:

[tex]\[
4x^2 - 21x + 5 = 4x^2 - 20x - x + 5
\][/tex]

- Group the terms and factor by grouping:

[tex]\[
= 4x(x - 5) - 1(x - 5)
\][/tex]

- Factor out the common factor [tex]\((x - 5)\)[/tex]:

[tex]\[
= (4x - 1)(x - 5)
\][/tex]

### Step 4: Write the complete factorization.
Thus, the complete factorization of [tex]\(f(x)\)[/tex] is:

[tex]\[
f(x) = (x + 9)(x - 5)(4x - 1)
\][/tex]

So, the polynomial [tex]\(f(x) = 4x^3 + 15x^2 - 184x + 45\)[/tex] factors completely as [tex]\((x + 9)(x - 5)(4x - 1)\)[/tex].