ok im almost done with this math equation but the end of it needs you to factor it and i cant seem to figure it out..anyone know?
w² +3w+8
2006-07-28 18:14:07 · 13 answers · asked by aliCia<3syOu 2 in Science & Mathematics ➔ Mathematics
First you find the roots of this equation:
(-3+SQRT(3^2-4*1*8))/2*1=-1.5+SQRT(5.75)i,
(-3-SQRT(3^2-4*1*8))/2*1=-1.5-SQRT(5.75)i;
then you negate both and add w:
-(-1.5+SQRT(5.75)i)=1.5-SQRT(5.75)i,
-(-1.5-SQRT(5.75)i)=1.5+SQRT(5.75)i,
w+(1.5-SQRT(5.75)i)=w-1.5-SQRT(5.75)i,
w+(1.5+SQRT(5.75)i)=w+1.5-SQRT(5.75)i,
then multiply:
w^2+3w+8=(w-1.5-SQRT(5.75)i)(w+1.5-SQRT(5.75)i)
The presence of complex numbers in the roots already told you that this equation is prime, but if you want the complex factorisation, this is it.
2006-07-28 18:28:15 · answer #1 · answered by angyansheng65537 2 · 0⤊ 0⤋
here given = 3 w^2 -10w - 8 = 3w^2 -12w +2w -8 , destroy -10w into 2 aspects -12 +2 and additionally -12*2 =24 = 3w (w - 4) +2 (w - 4) now take person-friendly 3w from 1st and 2d and +2 from third and 4th we get = (w - 4) (3w + 2) as a result w = 4, -2/3
2016-11-03 05:57:14 · answer #2 · answered by ? 4 · 0⤊ 0⤋
w^2 + 3w + 8
Can't be factored and here's why
(w + 8)(w + 1) = w^2 + 9w + 8
(w + 4)(w + 2) = w^2 + 6w + 8
2006-07-29 04:11:22 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
It has to begin with w standing alone for both factors. The other side is an integer. The integers must add up to 3 and multiply up to 8. I cant find any combination that can do that. This equation is unsolvable.
2006-07-28 18:23:21 · answer #4 · answered by eric l 6 · 0⤊ 0⤋
While answering the two numbers need to add up to 3 and their product must be 8, but using this we cant find any such numbers that will fit your equation, the equation is wrong
2006-07-28 18:27:21 · answer #5 · answered by Worst 2 · 0⤊ 0⤋
can not be factor
when a quadratic equation can not be factor you can use the quadratic formula to solve it.
x equals the opposite of B , plus or minus the square root of B squared minus 4AC, all divided by 2A
2006-07-28 19:38:54 · answer #6 · answered by ? 2 · 0⤊ 0⤋
It can be factorised mathematically to have answers in imaginery form. In real number there is no factor of the expression.
2006-07-28 21:16:46 · answer #7 · answered by sharanan 2 · 0⤊ 0⤋
The formula is
-b+sqrt (b^2-4ac)/4ab for the eqn ax^2+bx+c
2006-07-28 21:12:04 · answer #8 · answered by Anonymous · 0⤊ 0⤋
try a method call "perfecting the square". it is an alternative form of the eqn since it cannot be factorised
2006-07-28 19:06:22 · answer #9 · answered by sp_jimmy 1 · 0⤊ 0⤋
it's not posible...that's it. u just split the first w's in 2.
(w+ )(w+ )
then u try to find wat 2 #s mult. to the 3rd # but added together to get the mid. #. so wat 2 #s mult. up to 8 but adds together to get 3.
well...2&4 doesn't add to 3. neither does 1&8. so it's impossible.
2006-07-28 18:43:39 · answer #10 · answered by Carmen 3 · 0⤊ 0⤋