Question

simplify the following expression.
$(x^2 - 6x + 1)(3x - 4)$
$?x^3 + \square x^2 + \square x + \square$

Explanation:

Step1: Distribute \(x^2\)

Multiply \(x^2\) by each term in \((3x - 4)\): \(x^2 \cdot 3x = 3x^3\), \(x^2 \cdot (-4) = -4x^2\)

Step2: Distribute \(-6x\)

Multiply \(-6x\) by each term in \((3x - 4)\): \(-6x \cdot 3x = -18x^2\), \(-6x \cdot (-4) = 24x\)

Step3: Distribute \(1\)

Multiply \(1\) by each term in \((3x - 4)\): \(1 \cdot 3x = 3x\), \(1 \cdot (-4) = -4\)

Step4: Combine like terms

• \(x^3\) term: \(3x^3\)
• \(x^2\) terms: \(-4x^2 - 18x^2 = -22x^2\)
• \(x\) terms: \(24x + 3x = 27x\)
• Constant term: \(-4\)

Answer:

\(3x^3 - 22x^2 + 27x - 4\)

So the coefficients are: \(3\) (for \(x^3\)), \(-22\) (for \(x^2\)), \(27\) (for \(x\)), and \(-4\) (constant).