Question

How do you factor `x^3 + 10x^2 + 33x + 34`?

Answer 1

`x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)`

Explanation:

As putting `x=-2`, we get `x^3+10x^2+33x+34=(-2)^3+10xx2^2+33xx(-2)+34=-8+40-66+34=0`. Hence, `-2` is a zero of the function and `(x+2)` is a factor of `x^3+10x^2+33x+34`.

Now dividing the function by `(x+2)`, as `x^3+10x^2+33x+34=^=x^2(x+2)+8x(x+2)+17(x+2)=(x^2+8x+17)(x+2)`, we get `x^2+8x+17`.

Now discriminant of `x^2+8x+17` is `8^2-4*1*17=64-68=-4` is negative, as such it cannot be factorized in real factors.

And hence factors of the function are

`x^3+10x^2+33x+34=^=(x^2+8x+17)(x+2)`