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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!

 Jun 8, 2019

Best Answer 

 #1
avatar+8215 
+2

\(\ \phantom{=\quad}\frac43x^2+4x+1\)

                                                   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.

 

\(abc\ =\ (\frac43)(\frac32)(-2)\ =\ (2)(-2)\ =\ -4\) _

 Jun 8, 2019
 #1
avatar+8215 
+2
Best Answer

\(\ \phantom{=\quad}\frac43x^2+4x+1\)

                                                   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.

 

\(abc\ =\ (\frac43)(\frac32)(-2)\ =\ (2)(-2)\ =\ -4\) _

hectictar Jun 8, 2019

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