+379 The quadratic $\frac43x^2+4x+1$ can be written in the form a(x+b)^2+c, where a,b , and c are constants. What is abc? Give your answer in simplest form.
Thank you!
+9468 \(\ \phantom{=\quad}\frac43x^2+4x+1\)
Factor \(\frac43\) out of all the terms.
\(=\quad\frac43(x^2+3x+\frac34)\)
Add \(\frac94\) and subtract \(\frac94\) to complete the square inside the parenthesees.
\(=\quad\frac43(x^2+3x+\frac{9}{4}-\frac{9}{4}+\frac34)\)
Factor \(x^2+3x+\frac94\) as a perfect square trinomial.
\(=\quad\frac43(\ (x+\frac32)^2-\frac{9}{4}+\frac34)\)
Combine \(-\frac94\) and \(+\frac34\)
\(=\quad\frac43(\ (x+\frac32)^2-\frac{6}{4})\)
Distribute the \(\frac43\)
\(=\quad\frac43(x+\frac32)^2-\frac43(\frac{6}{4})\)
And \(\frac43(\frac64)\ =\ 2\)
\(=\quad\frac43(x+\frac32)^2-2\)
Now the expression is in the form a(x + b)2 + c where a, b, and c are constants.
\(abc\ =\ (\frac43)(\frac32)(-2)\ =\ (2)(-2)\ =\ -4\) _
+9468 \(\ \phantom{=\quad}\frac43x^2+4x+1\)
Factor \(\frac43\) out of all the terms.
\(=\quad\frac43(x^2+3x+\frac34)\)
Add \(\frac94\) and subtract \(\frac94\) to complete the square inside the parenthesees.
\(=\quad\frac43(x^2+3x+\frac{9}{4}-\frac{9}{4}+\frac34)\)
Factor \(x^2+3x+\frac94\) as a perfect square trinomial.
\(=\quad\frac43(\ (x+\frac32)^2-\frac{9}{4}+\frac34)\)
Combine \(-\frac94\) and \(+\frac34\)
\(=\quad\frac43(\ (x+\frac32)^2-\frac{6}{4})\)
Distribute the \(\frac43\)
\(=\quad\frac43(x+\frac32)^2-\frac43(\frac{6}{4})\)
And \(\frac43(\frac64)\ =\ 2\)
\(=\quad\frac43(x+\frac32)^2-2\)
Now the expression is in the form a(x + b)2 + c where a, b, and c are constants.
\(abc\ =\ (\frac43)(\frac32)(-2)\ =\ (2)(-2)\ =\ -4\) _
