Question

6.4 trinomials where a does not equal 1: problem 16 (1 point) factor the polynomial 20x² + 51x + 28. your answer can be written as (5x + b)(cx + d) with b, c, and d- integers where b equals: and c equals: and d equals: note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor

Explanation:

Step1: Multiply the coefficient of \(x^{2}\) and the constant term

For the polynomial \(20x^{2}+51x + 28\), \(a = 20\), \(b=51\), \(c = 28\). Calculate \(a\times c=20\times28 = 560\).

Step2: Find two numbers that multiply to \(560\) and add up to \(51\)

We need to find \(m\) and \(n\) such that \(m\times n=560\) and \(m + n=51\). The numbers are \(35\) and \(16\) (\(35\times16 = 560\) and \(35 + 16=51\)).

Step3: Rewrite the middle - term

Rewrite \(20x^{2}+51x + 28\) as \(20x^{2}+35x+16x + 28\).

Step4: Group the terms

\((20x^{2}+35x)+(16x + 28)\).

Step5: Factor out the common factors from each group

Factor out \(5x\) from the first group: \(5x(4x + 7)\). Factor out \(4\) from the second group: \(4(4x + 7)\).

Step6: Factor the polynomial

The polynomial becomes \((5x + 4)(4x+7)\)

Answer:

\(B = 4\), \(C = 4\), \(D = 7\)