How to factor 2x^2+5x-3???

- anonymous

How to factor 2x^2+5x-3???

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- jim_thompson5910

Have you learned the quadratic formula?

- anonymous

Yes.
Would this be applied here?

- anonymous

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- jim_thompson5910

yes the best way to factor in my opinion is to first find the roots using the quadratic formula. What roots do you get?

- anonymous

25-(24)=1

- jim_thompson5910

ok that looks like the value of b^2 - 4ac

- jim_thompson5910

notice that since the discriminant is equal to 1, which is a positive perfect square, this means we can factor

- anonymous

I see... so how do we proceed?

- anonymous

I attempted to utilize the AC method when going about factoring the trinomial but ended up with an incorrect result.

- jim_thompson5910

so do you have this written down?
\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-5\pm\sqrt{5^2-4*2*(-3)}}{2*2}\]
\[\Large x = \frac{-5\pm\sqrt{49}}{4}\]
notice how 49 is under the root and NOT 1

- anonymous

Yes; would we proceed by simplifying the division and the square root? -5+/-7/4= 1/2 or -3.

- jim_thompson5910

correct, the two roots are 1/2 and -3

- jim_thompson5910

if x = 1/2, then we can multiply both sides by 2 to get 2x = 1
then subtract the 1 from both sides to get 2x - 1 = 0
agreed?

- anonymous

Certainly.

- jim_thompson5910

then if x = -3, we can add 3 to both sides to get x+3 = 0

- anonymous

I see where you're going now, by discovering the roots, it's much easier to find the factors.

- jim_thompson5910

so the two roots
x = 1/2 or x = -3
turns into these two equations
2x-1 = 0 or x+3 = 0

- jim_thompson5910

yes, we then use the zero product property to go from `2x-1 = 0 or x+3 = 0` to `(2x-1)(x+3)=0`

- jim_thompson5910

the last thing to do is to expand out `(2x-1)(x+3)` and see if you get `2x^2+5x-3`

- anonymous

While this certainly is an easier method, I don't know if in application I will be able to know to multiply the root, x=1/2, by 2 on both sides. What tells us we need to do this?

- jim_thompson5910

the goal is to get each root to turn into an equation where the right side is 0
so we can use the fact that `A = 0 or B = 0` turns into `A*B = 0`

- jim_thompson5910

you dont' have to multiply by 2, but I don't like fractions which is why I try to get rid of them as soon as possible

- jim_thompson5910

you could easily have
x - (1/2) = 0 and x+3 = 0
turn into
(x - (1/2))*(x+3) = 0
but that's a bit more complicated than it needs to be

- anonymous

I guess I would remember to do it as the factors typically need to be written with whole numbers.

- jim_thompson5910

yeah that's a good general rule to have so it's as simple as possible

- anonymous

Thank you so much for your thorough response!

- jim_thompson5910

you're welcome

Looking for something else?

Not the answer you are looking for? Search for more explanations.