Question

Can you factor 2x^4-5x^3+53x^2-125x+75 into 2 quadratic expressions?

Answer 1

\[2x ^{4}-5x ^{3}+53x^2-125x+75\]

Answer 2

@SolomonZelman

Answer 3

Well, reducing fully, \(\normalsize\color{blue}{ (x-1)(2x+3)(x^2+25)\LARGE\color{white}{ \rm │ }}\) .

Answer 4

So yes, you will get \(\normalsize\color{blue}{ (x^2+x-3)(x^2+25)~.\LARGE\color{white}{ \rm │ }}\)

Answer 5

Ok thanks, and then to find the zeros, I use (x-1)(2x-3)(x+5)(x-5)? @SolomonZelman

Answer 6

No, it is x\(\normalsize\color{blue}{ +}\)25, not -.

Answer 7

x^2+25=0 --> x=±5?

Answer 8

You will get

\(\normalsize\color{blue}{ (x-1)(2x-3)(x^2+25)=0\LARGE\color{white}{ \rm │ }}\)
─────────────────
\(\normalsize\color{blue}{ (x-1)=0~~~~⇒~~~~x=1\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{ (2x-3)=0~~~~⇒~~~~x=3/2\LARGE\color{white}{ \rm │ }}\)
─────────────────
or, \(\normalsize\color{blue}{x^2+25=0\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{x^2=-25\LARGE\color{white}{ \rm │ }}\)
\(\normalsize\color{blue}{x=±5i\LARGE\color{white}{ \rm │ }}\)

Answer 9

Right, I see, the 25 would be negative when you move it over so you have to use an imaginary number to get the square root.

Answer 10

Yes, exactly :)

Answer 11

OK, thanks :D

Answer 12

Anytime:)