Question

Find the roots of x^3-2x^2+10x+136=0

Answer 1

\[x ^{3}-2x ^{2}+10x+136=0\]

Answer 2

\(\normalsize\color{black}{ (x+4)(x^2-6x+34)=0\LARGE\color{white}{ \rm │ }}\)

Answer 3

I factored it, can you do it from here ?

Answer 4

Solve for x:
\[x^3-2 x^2+10 x+136 = 0 \]
Factor the left hand side.
The left hand side factors into a product with two terms:
\[(x+4) (x^2-6 x+34) = 0 \]
Solve each term in the product separately.
Split into two equations:
\[x+4 = 0 \] or \[x^2-6 x+34 = 0 \]
Look at the first equation: Solve for x.
Subtract 4 from both sides:
\[x = -4 \] or \[x^2-6 x+34 = 0 \]
Look at the second equation: Solve the quadratic equation by completing the square.
Subtract 34 from both sides:
\[x = -4 \] or \[x^2-6 x = -34 \]
Take one half of the coefficient of x and square it, then add it to both sides.
Add 9 to both sides:
\[x = -4 \] or \[x^2-6 x+9 = -25 \]
Factor the left hand side.
Write the left hand side as a square:
\[x = -4 \] or \[(x-3)^2 = -25 \]
Eliminate the exponent on the left hand side.
Take the square root of both sides:
\[x = -4 \] or \[x-3 = 5 i \] or \[x-3 = -5 i \]
Look at the second equation: Solve for x.
Add 3 to both sides:
\[x = -4 \] or \[x = 3+5 i \] or \[x-3 = -5 i \]
Look at the third equation: Solve for x.
Add 3 to both sides:

Answer 5

99% of the work:)

Answer 6

I got suspended for this number of times

Answer 7

Pardon?

Answer 8

I don't mind

Answer 9

I mean, \(\LARGE\color{black}{ \rm I\LARGE\color{white}{ \rm │ }}\)don't mind.

Answer 10

I am also a fan of completing the square, btw.

Answer 11

Thanks :D

Answer 12

No problem, what'd you get?

Answer 13

x=-5i+3, x=5i+3, x=-4

Answer 14

Or you could do x=3+5i, x=3-5i for the first 2.