Question

question
solve the quadratic by factoring.
$3x^2 + 16x + 1 = 9x - 3$
answer attempt 1 out of 10
$oplus$ additional solution $ominus$ no solution
$x = \square$
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Explanation:

Step1: Rearrange the equation to standard form

First, we need to move all terms to the left side of the equation to get it in the standard quadratic form \(ax^2 + bx + c = 0\).
Starting with \(3x^2 + 16x + 1 = 9x - 3\), we subtract \(9x\) and add \(3\) to both sides:
\(3x^2 + 16x - 9x + 1 + 3 = 0\)
Simplifying the like terms:
\(3x^2 + 7x + 4 = 0\)

Step2: Factor the quadratic equation

We need to find two numbers that multiply to \(3\times4 = 12\) and add up to \(7\). The numbers \(3\) and \(4\) work because \(3\times4 = 12\) and \(3 + 4 = 7\).
We rewrite the middle term using these numbers:
\(3x^2 + 3x + 4x + 4 = 0\)
Now, we factor by grouping:
Group the first two terms and the last two terms:
\((3x^2 + 3x) + (4x + 4) = 0\)
Factor out the greatest common factor from each group:
\(3x(x + 1) + 4(x + 1) = 0\)
Now, we can factor out \((x + 1)\) from both terms:
\((3x + 4)(x + 1) = 0\)

Step3: Solve for \(x\) using the zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\).
So, we set each factor equal to zero:

1. \(3x+4 = 0\)

Subtract \(4\) from both sides: \(3x=-4\)
Divide both sides by \(3\): \(x =-\frac{4}{3}\)

1. \(x + 1 = 0\)

Subtract \(1\) from both sides: \(x=-1\)

Answer:

\(x=-1\) or \(x =-\frac{4}{3}\)