+0

-1
663
1
+395

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

#1
+9180
+3

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

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
+9180
+3

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

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