Question

6.4 trinomials where a does not equal 1: (1 point) factor: $8x^2 + 18x + 7 = \square$ preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor page gene webwork © 1996-202

Explanation:

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

For the trinomial \(8x^{2}+18x + 7\), \(a = 8\), \(b=18\), \(c = 7\). Calculate \(a\times c=8\times7 = 56\)

Step2: Find two numbers that multiply to \(a\times c\) and add up to \(b\)

We need to find two numbers \(m\) and \(n\) such that \(m\times n=56\) and \(m + n=18\). The numbers are \(14\) and \(4\) since \(14\times4=56\) and \(14 + 4=18\)

Step3: Rewrite the middle - term

Rewrite \(8x^{2}+18x + 7\) as \(8x^{2}+14x+4x + 7\)

Step4: Group the terms and factor by grouping

\((8x^{2}+14x)+(4x + 7)=2x(4x + 7)+1(4x + 7)\)

Step5: Factor out the common binomial factor

\((2x + 1)(4x+7)\)

Answer:

\((2x + 1)(4x + 7)\)