To express the polynomial [tex]\( g(x) = x^3 + 5x^2 + 2x - 12 \)[/tex] as a product of linear factors, we need to follow these steps:
1. Identify a Zero:
We are given that [tex]\(-3\)[/tex] is a zero of [tex]\(g(x)\)[/tex]. This means that [tex]\( g(-3) = 0 \)[/tex].
2. Construct the Factorization Based on the Zero:
Since [tex]\(-3\)[/tex] is a zero, [tex]\( x + 3 \)[/tex] is a factor of [tex]\( g(x) \)[/tex]. Our goal is to perform polynomial division to find the other factors.
3. Polynomial Division:
We will divide [tex]\( g(x) \)[/tex] by [tex]\( x + 3 \)[/tex].
To do this, simulate the polynomial long division:
- Divide the leading term [tex]\( x^3 \)[/tex] by [tex]\( x \)[/tex], which gives [tex]\( x^2 \)[/tex].
- Multiply [tex]\( x + 3 \)[/tex] by [tex]\( x^2 \)[/tex], which gives [tex]\( x^3 + 3x^2 \)[/tex].
- Subtract [tex]\( x^3 + 3x^2 \)[/tex] from [tex]\( g(x) \)[/tex] to get the new polynomial:
[tex]\[ g(x) - (x^3 + 3x^2) = (x^3 + 5x^2 + 2x - 12) - (x^3 + 3x^2) = 2x^2 + 2x - 12 \][/tex]
- Next, divide the leading term of the new polynomial [tex]\( 2x^2 \)[/tex] by [tex]\( x \)[/tex], giving [tex]\( 2x \)[/tex].
- Multiply [tex]\( x + 3 \)[/tex] by [tex]\( 2x \)[/tex], which gives [tex]\( 2x^2 + 6x \)[/tex].
- Subtract [tex]\( 2x^2 + 6x \)[/tex] from the new polynomial:
[tex]\[ 2x^2 + 2x - 12 - (2x^2 + 6x) = -4x - 12 \][/tex]
- Now, divide the leading term of this new polynomial [tex]\( -4x \)[/tex] by [tex]\( x \)[/tex], giving [tex]\( -4 \)[/tex].
- Multiply [tex]\( x + 3 \)[/tex] by [tex]\( -4 \)[/tex], which gives [tex]\( -4x - 12 \)[/tex].
- Subtract [tex]\( -4x - 12 \)[/tex] from the current polynomial:
[tex]\[ -4x - 12 - (-4x - 12) = 0 \][/tex]
The quotient from the division is [tex]\( x^2 + 2x - 4 \)[/tex]. Therefore, we can write:
[tex]\[ g(x) = (x + 3)(x^2 + 2x - 4) \][/tex]
4. Factor the Quadratic Polynomial:
Next, we factor the quadratic polynomial [tex]\( x^2 + 2x - 4 \)[/tex].
To factorize [tex]\( x^2 + 2x - 4 \)[/tex], we need the roots of the quadratic equation [tex]\( x^2 + 2x - 4 = 0 \)[/tex]. Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -4 \)[/tex]:
[tex]\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 16}}{2} = \frac{-2 \pm \sqrt{20}}{2} = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5}
\][/tex]
The roots are [tex]\( x = -1 + \sqrt{5} \)[/tex] and [tex]\( x = -1 - \sqrt{5} \)[/tex]. Therefore, the quadratic can be factored as:
[tex]\[
x^2 + 2x - 4 = (x - (-1 + \sqrt{5}))(x - (-1 - \sqrt{5})) = (x + 1 - \sqrt{5})(x + 1 + \sqrt{5})
\][/tex]
5. Combine All Factors:
Finally, we can express [tex]\( g(x) \)[/tex] as a product of linear factors:
[tex]\[
g(x) = (x + 3)(x + 1 - \sqrt{5})(x + 1 + \sqrt{5})
\][/tex]
Therefore, the polynomial [tex]\( g(x) \)[/tex] expressed as a product of linear factors is:
[tex]\[
g(x) = (x + 3)(x + 1 - \sqrt{5})(x + 1 + \sqrt{5})
\][/tex]
