Questions: Given that x-6 is a factor of P(x)=x^3-8 x^2+9 x+18, factor the polynomial in terms of linear factors. P(x)=

Transcript text: Given that $x-6$ is a factor of $P(x)=x^{3}-8 x^{2}+9 x+18$, factor the polynomial in terms of linear factors. \[ P(x)= \]

Solution

Solution Steps

Solution Approach

Since $x-6$ is a factor of $P(x)$, use polynomial division or synthetic division to divide $P(x)$ by $x-6$.
The quotient will be a quadratic polynomial.
Factor the resulting quadratic polynomial into linear factors.
Combine all the linear factors to express $P(x)$ in terms of linear factors.

Step 1: Identify the Factor

Given that $x - 6$ is a factor of the polynomial $P(x) = x^3 - 8x^2 + 9x + 18$, we can express $P(x)$ as: \[ P(x) = (x - 6)Q(x) \] where $Q(x)$ is a quadratic polynomial.

Step 2: Perform Polynomial Division

By dividing $P(x)$ by $x - 6$, we find the quotient: \[ Q(x) = x^2 - 2x - 3 \]

Step 3: Factor the Quadratic Polynomial

Next, we factor the quadratic polynomial $Q(x)$: \[ Q(x) = (x - 3)(x + 1) \]

Step 4: Combine All Factors

Now, we can express the original polynomial $P(x)$ in terms of its linear factors: \[ P(x) = (x - 6)(x - 3)(x + 1) \]