Question

i need help on factoring 2t^2+5t-3

Answer 1

'Which methods of factoring quadratics have you used in the past?

1) completing the square
2) factoring by grouping
3) synthetic division
4) Rational Root Theorem

Try one or more of these.

Answer 2

can you help me with the factoring by grouping?

Answer 3

We could certainly try that. Unfortunately, factoring by grouping would be difficult to apply here, although we might be able to verify factors found in other ways.

Answer 4

I just named several methods of factoring quadratics, not meaning to recommend any of them by the order i which I presented them.

Answer 5

i would just like to know how to do one of them please

Answer 6

Try this: Take the first coefficient (2) and the last coefficient (-3) and multiply them together. What are factors of the result?

(2)(-3)=?

Answer 7

-6

Answer 8

So now what do i do @mathmale

Answer 9

List all of the possible factors of -6: \[\pm 6;~\pm 3;~\pm 2;~\pm 1\]

Answer 10

If you know synthetic div, I'd strongly recommend that you use it to determine whether or not any of those 8 possible roots are indeed solutions of the given quadratic. If y ou don't know synth. div., there are other methods, such as long division.

Answer 11

i am quite lost

Answer 12

All right. Let's revert to the quadratic formula (which I should have listed earlier). Heard of and/or used that?

Answer 13

My teacher was going to explain that later this week but i dont mind giving it a try

Answer 14

\[If~ ax^2+bx+c=0,~x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 15

You are given the quadratic equation 2x^2 + 5x - 3 = 0, for which the coefficients a, b and c are 2, 5 and -3 respectively.

Try putting those values into the formula for x and calculating x that way.

Answer 16

so would my equation be x=-5+/- the square root of 5^2-4 over 2(2)

Answer 17

Good start. Please, could you put that into a nicer mathematical format (e. g., using a horiz line instead of "over")?

Simplify 5^2 - 4.

Answer 18

Well i would if i could

Answer 19

so would my equation be x=-5+/- the square root of 5^2-4 over 2(2)\[x=\frac{ -5\pm \sqrt{5^2-4} }{ 2(2) }\]

Answer 20

Check your "b^2-4ac." I think you've forgotten your "a" and "c".

Answer 21

b=5; a=2; and c=-3.

What is the "determinant," b^2 - 4ac?

( )^2 - 4 ( )( )

Answer 22

so is this what it should look like

Answer 23

Were you kicked off the 'Net or did you log off? If you log off in mid-stream, please let me know your intentions beforehand. I really wanted to help you.

Answer 24

Very good. What is 5^2? What is 4(2)(-3)?

Answer 25

OpenStudy first told me you were "offline" and then that you were "just looking around."

Answer 26

What is the value of 5^2 - 4(2)(-3)?

Answer 27

i was online and i got the app

Answer 28

I got 1

Answer 29

5^2 = 25

-4 (2) (-3) = ??

Answer 30

Yes, 25-24 =1.

Answer 31

So your two roots are:\[x=\frac{ -5\pm \sqrt{1} }{ 4 }\]

Answer 32

sorry