Question

Can `y=2x^2 – 3x – 5` be factored? If so what are the factors ?

Answer 1

`x=5/2,-1`

Explanation:

`2x^2-3x-5`

=`(2x-5)(x+1)`

`x=5/2,-1`

Answer 2

`(x+1)(2x-5)`

Explanation:

`"for a quadratic in "color(blue)"standard form"`

`•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0`

`"consider the factors of the product ac which sum to b"`

`"here "a=2,b=-3" and "c=-5`

`rArrac=2xx-5=-10" and - 5, + 2 sum to - 3"`

`color(blue)"split the middle term using these factors"`

`rArry=2x^2+2x-5x-5larrcolor(blue)"factor by grouping"`

`color(white)(y)=color(red)(2x)(x+1)color(red)(-5)(x+1)`

`"take out a "color(blue)"common factor "(x+1)`

`=(x+1)(color(red)(2x-5))`

`rArr2x^2-3x-5=(x+1)(2x-5)`

Answer 3

`y=(2x-5)(x+1)`

Explanation:

Can the trinomial be factored? `y = ax^2 +bx+c`

Test it using `" "b^2 -4ac" "` first.
If the answer is a perfect square then you can find factors. This means you do not waste time by trying to find factors if there are none!

`(-3)^2 -4(2)(-5) = 9+40 =49" "larr = 7^2`

Find factors of `2 and 5` whose products differ by `3`

Both `2 and 5` are prime numbers so there are only two combinations to try:

`(2xx5) - (1xx1) = 10-1=9`

`(1xx5)-(2xx1) =5-2=3`

`y=(2x-5)(x+1)`