i also need help with 12x^2+28x-24 i get that you can factor out a 4 giving you 3x^2+7x-6 so would it be (3x+7)(x-1)but when i multiply i get something that dose not work any ideas?
You don't factor x^2 + 3x + 14. Not with real numbers, anyway.
To factor 3x^2 + 7x - 8, you need radicals. Complete the square by:
3x^2 + 7x - 8 = 3 [x^2 + 7x/3 - 8/3]
= 3 [x^2 + 7x/3 + 49/36 - 49/36 - 8/3]
= 3[(x + 7/6)^2 - 49/36 - 96/36]
= (3/36)[(6x + 7)^2 - 145]
Now, that quantity in [] brackets is a difference of squares, which factors:
= (1/12) [(6x + 7) + √145] [(6x + 7 - √145]
That √145 has no square factors, so that's about as simple as you're going to get it. Multiply in the factor of 4 that leading fraction becomes 1/3.
x^2+3x+14 cannot be factored so you have to use the quadratic equation. You end up with a complex number solution.
You factored 3x^2+7x-6 wrong. Try (3x-2)(x+3).
Comments
You don't factor x^2 + 3x + 14. Not with real numbers, anyway.
To factor 3x^2 + 7x - 8, you need radicals. Complete the square by:
3x^2 + 7x - 8 = 3 [x^2 + 7x/3 - 8/3]
= 3 [x^2 + 7x/3 + 49/36 - 49/36 - 8/3]
= 3[(x + 7/6)^2 - 49/36 - 96/36]
= (3/36)[(6x + 7)^2 - 145]
Now, that quantity in [] brackets is a difference of squares, which factors:
= (1/12) [(6x + 7) + √145] [(6x + 7 - √145]
That √145 has no square factors, so that's about as simple as you're going to get it. Multiply in the factor of 4 that leading fraction becomes 1/3.
x^2+3x+14 cannot be factored so you have to use the quadratic equation. You end up with a complex number solution.
You factored 3x^2+7x-6 wrong. Try (3x-2)(x+3).
You don't factor x^2 + 3x + 14. Not with real numbers, anyway.