Question

Question #a3836

Answer 1

`-2(x+(11+sqrt(329))/4)(x-(sqrt(329)-11)/4)`, but we will say that it's not factorable.

Explanation:

`26-11x-2x^2=-2x^2-11x+26`

`-2x^2-11x+26` is not factorable.

The roots of `-2x^2-11x+26` are `x=-(11+sqrt(329))/4` and `x=(sqrt(329)-11)/4`.

`(x+(11+sqrt(329))/4)(x-(sqrt(329)-11)/4)=x^2+(11x)/2-13`

`:.-2(x^2+(11x)/2-13)=-2x^2-11x+26=-2(x+(11+sqrt(329))/4)(x-(sqrt(329)-11)/4)`

Answer 2

Q: Factorize: `y=-2x^2-11x+26`

A: `y=-2(x+11/4)^2+329/8`

Explanation:

Factorise by completing the square;
`y=-2x^2-11x+26`
1) Convert to a monic quadratic function by dividing both sides of the equality by `-2`
`y/(-2)=(-2x^2-11x+26)/(-2)`

`-y/2=x^2+(11x)/2-13`

2) Add 13 to both sides
`-y/2+13=x^2+(11x)/2`

3) Add the square of half of the `x` term `11/2` (which is `11/4`) to both sides
`-y/2+13+(11/4)^2=x^2+(11x)/2+(11/4)^2`

4) Combine the like terms on the left of the equality
`-y/2+329/16=x^2+(11x)/2+(11/4)^2`

5) What we have been trying to do is create a perfect square on the right hand side of the equality. Now it is time to factorise that perfect square.

`-y/2+329/16=(x+11/4)^2`

6) Now we are going to move all of the stuff on the left of the equality back to the right (barring the `y` variable of course)

`-y/2=(x+11/4)^2-329/16`

`y=-2[(x+11/4)^2-329/16]`

7) Multiplying the `-2` into the brackets will give us our final answer

`y=-2(x+11/4)^2+329/8`

I hope that helps :)

Harold