To decompose the polynomial \((x^2 - 7x + 6)(x^2 + 3x - 18)\) into linear factors, we need to factor each quadratic expression separately.
Step 1: Factor \(x^2 - 7x + 6\)
We look for two numbers that multiply to \(6\) (the constant term) and add to \(-7\) (the coefficient of \(x\)). The numbers \(-1\) and \(-6\) satisfy this because: \[ -1 \times -6 = 6 \quad \text{and} \quad -1 + -6 = -7 \]
Thus, we can factor \(x^2 - 7x + 6\) as: \[ (x - 1)(x - 6) \]
Step 2: Factor \(x^2 + 3x - 18\)
Now we look for two numbers that multiply to \(-18\) (the constant term) and add to \(3\) (the coefficient of \(x\)). The numbers \(6\) and \(-3\) satisfy this because: \[ 6 \times -3 = -18 \quad \text{and} \quad 6 + -3 = 3 \]
Thus, we can factor \(x^2 + 3x - 18\) as: \[ (x + 6)(x - 3) \]
Step 3: Combine the factors
Now, we can combine all these factors together:
\[ (x^2 - 7x + 6)(x^2 + 3x - 18) = (x - 1)(x - 6)(x + 6)(x - 3) \]
Therefore, the polynomial \((x^2 - 7x + 6)(x^2 + 3x - 18)\) can be decomposed into its linear factors as: \[ (x - 1)(x - 6)(x + 6)(x - 3) \]
1 answer