Question

Please help. WILL MEDAL! @jim_thompson5910 @Erak @Mehek14 @triciaal @hartnn @johnweldon1993 @satellite73

Answer 1

please don't mass tag especially when the question is not posted

Answer 2

Factor \[4x ^{2} + 8x + 28\] over the set of complex numbers.

Answer 3

\[4(x ^{2}+2x+7)\]

Answer 4

This is where I get stuck.

Answer 5

4x^2+8x+28 = 4*(x^2+2x+7)

now use the quadratic formula to solve x^2+2x+7 = 0

Answer 6

I know that I have to solve it with imaginary numbers

Answer 7

\[x = -1 \pm 6i\]

Answer 8

\[4(x+1-6i)(x+1+6i)\]

Answer 9

Thank you so much!

Answer 10

incorrect. You solved x^2+2x+7 = 0 incorrectly

Answer 11

What do you mean?

Answer 12

you lost a root somewhere

Answer 13

\[6i = \sqrt{-6}\]

Answer 14

\[\Large x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-2\pm \sqrt{2^2-4*1*7}}{2*1}\]
\[\Large x = \frac{-2\pm \sqrt{-24}}{2}\]
\[\Large x = \frac{-2\pm \sqrt{-1*4*6}}{2}\]
\[\Large x = \frac{-2\pm \sqrt{-1}*\sqrt{4}*\sqrt{6}}{2}\]
\[\Large x = \frac{-2\pm 2i*\sqrt{6}}{2}\]
\[\Large x = -1\pm i\sqrt{6}\]


`6i` is NOT the same as `i*sqrt(6)`

Answer 15

So the full factorization would be

\[\Large 4x^2+8x+28 = 4(x^2+2x+7)\]
\[\Large 4x^2+8x+28 = 4(x - (-1+i\sqrt{6}))(x - (-1-i\sqrt{6}))\]
\[\Large 4x^2+8x+28 = 4(x +1-i\sqrt{6})(x+1+i\sqrt{6})\]