(4 x + 3)(-2 x - 5)
2006-10-10 15:01:06 · 4 answers · asked by Korey's Mamacita 1 in Science & Mathematics ➔ Mathematics
Multiply everything in one term by everything in the other term.
If you have (A+B)(C+D), it expands to AC+AD+BC+BD. For your case, just be careful with negatives.
This works for more as well. Somewhat combinatorially, you can pick one term from each set of parenthesis, and make that one term of your expansion.
That is, for (A+B+C)(D+E+F)(G+H+I), ADG, ADH, ADI, BEG, and CFG are all terms in your expansion (though not all of them.)
2006-10-10 15:06:44 · answer #1 · answered by zex20913 5 · 0⤊ 0⤋
(4x + 3)(-2x - 5)
(4x * -2x) + (4x * -5) + (3 * -2x) + (3 * -5)
-8x^2 - 20x - 6x - 15
-8x^2 - 26x - 15
-------------------------------
or
(4x + 3)(-2x - 5)
-(4x + 3)(2x + 5)
-(8x^2 + 20x + 6x + 15)
-(8x^2 + 26x + 15)
-8x^2 - 26x - 15
2006-10-10 22:35:45 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
-8x ^2 - 26x - 15
2006-10-10 22:07:04 · answer #3 · answered by lovinthemonster92 1 · 0⤊ 0⤋
(4x)(-2x) - (4x)(5) + (3)(-2x) - (3)(5)
-8x2 - 20x + -6x - 15
-8x2 - 26x - 15
2006-10-10 22:10:41 · answer #4 · answered by First L 1 · 0⤊ 0⤋