2006-11-20 05:21:53 · 16 answers · asked by ME 3 in Science & Mathematics ➔ Mathematics
How did you get 16 from 10x+6 arent they suppose to be the same in order to add them? for example 10x+6x? Thank you all for your help youve all helped alot
2006-11-20 05:30:10 · update #1
Dear friend,
Your product is
-5*(x^2) + 8*x -10*x + 16 =
-5*(x^2) - 2*x + 16
2006-11-20 05:57:25 · answer #1 · answered by Verbena 6 · 0⤊ 0⤋
FOIL = First, Outter, Inner, Last
(x+2)(-5x + 8) = -5x^2 + 8x - 10x +16
combine like terms, in this case the x terms
-5x^2 - 2x + 16
2006-11-20 05:39:55 · answer #2 · answered by Pony 2 · 0⤊ 0⤋
Fast outer inner last
=(X+2)(-5X+8)
=-5x^2 + 8X - 10X + 16
=-5x^2 -2x + 16
2006-11-20 05:35:50 · answer #3 · answered by Anonymous · 0⤊ 0⤋
(X+2)(-5X+8) Use FOIL
-5x^2+8x-10x+16
-5x^2-2x+16
2006-11-20 05:23:41 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
using foil is the way to do it
(x+2)(-5x+8) FOIL = First Outside Inside Last
(-5x^2+16-10x+8x)
(-5^2-2x+16)
then do the roots i hope thats how you do it!! lol
2006-11-20 05:47:40 · answer #5 · answered by shamrock_bhoy 1 · 0⤊ 0⤋
set both functions equal to zero separately
x+2=o
x=-2
-5x+8=0
-5x=-8
x=8/5
those are the two answers .
to multiply it out use FOIL method
(first, outer, inner, last)
-5x^2+8x-10x+16
-5x^2 -2x+16
to solve for x this way use the quadratic equation
x=-b+/- square root of b^2 -4ac
all over 2a
2006-11-20 05:27:16 · answer #6 · answered by Anonymous · 0⤊ 1⤋
{x+2} {- 5x+8}
-5x2[squared] +8x - 10x +16
-5x2 - 2x +16
2006-11-20 09:52:52 · answer #7 · answered by Anonymous · 0⤊ 0⤋
=-5x^2+8x-10x+16
=-5x^2-2x+16
2006-11-20 05:26:01 · answer #8 · answered by openpsychy 6 · 0⤊ 0⤋
x * -5x plus
x * 8 plus
2 * -5x plus
2 * 8 equals
-5x^2 + 8x -10x + 16 =
-5x^2 -2x + 16
2006-11-20 05:28:30 · answer #9 · answered by hznfrst 6 · 0⤊ 0⤋
is that meant to = 0? x^2+8/5x-9/25=0 whilst coping with fractions i could in basic terms use the quadratic formula. x = (-b +/- sqrt(b^2 - 4ac)) / 2a x = one million/5 , -9/5 (x-one million/5)(x+9/5)=0
2016-12-29 06:28:40 · answer #10 · answered by Anonymous · 0⤊ 0⤋