how do i simplify this?
2x+19x+24
2x-3x-9
this is a fraction and both 2x are squared.
2007-01-01 13:38:38 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
completly stuck on homework, need help!! thanks
2007-01-01 13:43:11 · update #1
2x^2 + 19X + 24
To factor this, we look for something that looks like (2x+a)(x+b)
so that ab = 24 and 2b+a = 19, and a and b are factors of 24.
If b = 8, this works (trial and error)
(2x+3)(x+8) = 2x^2 + 3x+16x+24
2x^2 - 3x - 9
In this case, it is going to look like either
(2x-a)(x+b) or (2x+a)(x-b)
factors of 9 are either 1 and 9 or 3 and 3
If we say (2x+3)(x-3) then we get 2x^2 +3x - 6x - 9, which works!
So now you have a fraction:
(2x+3)(x+8) divided by (2x+3)(x-3)
and the (2x+3) terms cancel, and you have (x+8)/(x-3)
2007-01-01 13:43:29 · answer #1 · answered by firefly 6 · 0⤊ 0⤋
What do you mean? If you mean that this is
(x^2 + 19x + 24) / (x^2 - 3x - 9), then you need to factorize the numerator and the denominator, and the cancel out if possible. Neither of these polynomials is factorable, so this fraction is already in simplest terms.
If you mean that this is (2x^2 + 19x + 24) / (2x^2 - 3x - 9), then you still factor, and this case they are both factorable as shown by the previous answerer. Then you can cancel out (2x + 3) from the numerator and the denominator, leaving you with (x + 8) / (x - 3).
2007-01-01 21:44:57 · answer #2 · answered by DavidK93 7 · 0⤊ 2⤋
You need to factor the numerator and denominator.
(2x² + 19x + 24)/(2x² - 3x - 9)
= (2x + 3)(x + 8)/{(2x + 3)(x - 3)}
= (x + 8)/(x - 3)
However the function is undefined at x = -3/2 and 3. This is because the denominator in the original function would be zero for those values of x, leaving the function undefined.
2007-01-01 21:54:39 · answer #3 · answered by Northstar 7 · 0⤊ 0⤋
(2x^2 +19x +24)/(2x^2-3x-9)
=[(2x+3)(x+8)]/[(2x+3)(x-3)]
= (x+8)/(x-3)
Notice that the factors 2x=3 cancel out allowing simplification.
2007-01-01 21:52:43 · answer #4 · answered by ironduke8159 7 · 0⤊ 0⤋
(2x^2+19x+24)/(2x^2-3x-9)
(2x+3)(x+8)/((2x+3)(x-3)) divide numerator & denominator by 2x+3
(x+8)/(x-3)
2007-01-01 22:38:35 · answer #5 · answered by yupchagee 7 · 0⤊ 0⤋
you need to factor both the numerator and denominator:
[(2x + 3)(x + 8 )]/[(2x +3)(x - 3)]
Then you can cancel the (2x + 3) term because it is in both the numerator and denominator. then you're left wih:
(x + 8)/(x-3)
2007-01-01 21:46:30 · answer #6 · answered by Marcella S 5 · 1⤊ 0⤋