Factoring Question (4x+4)(x+3)=2?

2008-09-17 11:52 am
I tried to factor: (4x+4)(x+3)=2.
First I factored out the left side to 4x^2+16x+12=2, and then subtracted the 2 to the left side to make the equation equal 0.
4x^2+16x+10=0. Then I took out the common multiple of 2 to get 2(2x^2+8x+5). Now I am lost.
更新1:

So what would the values of "x" be for the factored equation?

回答 (7)

2008-09-17 12:27 pm
✔ 最佳答案
(4x + 4)(x + 3) = 2
4x*x + 4*x + 4x*3 + 4*3 = 2
4x^2 + 4x + 12x + 12 = 2
4x^2 + 16x + 12 - 2 = 0
4x^2 + 16x + 10 = 0
(4x^2 + 16x + 10)/2 = 0/2
2x^2 + 8x + 5 = 0 (<== cannot be factored)
x = [-b ±√(b^2 - 4ac)]/2a (<== quadratic formula)

a = 2
b = 8
c = 5

x = [-8 ±√(64 - 40)]/4
x = [-8 ±√24]/4
x ≈ [-8 ±4.89]/4

x ≈ [-8 + 4.89]/4
x ≈ -3.11/4
x ≈ -0.7775

x ≈ [-8 - 4.89]/4
x ≈ -12.89/4
x ≈ -3.2225

∴ x ≈ -3.2225 , -0.7775
2008-09-17 7:03 pm
You have reached up to
2*(2x^2+8x+5) = 0
Divide both sides by 2.
2x^2+8x+5 = 0

This is a quadratic equation in x of the form
ax^2 + bx + c = 0
where a = 2, b = 8, c = 5
Discriminant D = b^2 - 4as = 8^2 - 4*2*5 = 64 - 40 = 24
The roots are
x = {-b +- sqrt(D)}/(2a)
= {-8 +- sqrt(24)}/(2*2)
= {-8 +- sqrt(4*6)}/4
= {-8 +- 2*sqrt(6)}/4
= -2 +- sqrt(6)/2
2008-09-17 6:58 pm
Looks to me like it's already factorized. And only an Algebric Expression can be factorized.
2008-09-17 6:57 pm

It is already in factored form. You don't factor equations,
you factor expressions.
2008-09-17 7:00 pm
like the first dude said
2008-09-17 6:59 pm
you don't need do it from this way
4x+4=4(x+1)
4(x+1)(x+3)=2
if you solve this question you cant solve it from factoring you must multiply and is 4x^2+16x+10=0
that your x1=-3.2247 x2=-0.7752
2008-09-17 7:11 pm
2(2x^2+8x+5) = 0 reduces to 2x^2+8x+5 = 0

To solve the quadtratic equation, ax²+bx+c=0

x= (-b+ √b²-4ac)/2a and (-b-√b²-4ac)/2a

Substituting a=2, b=8, c=5 in the above equations, we get,

x= 2.9 and -6.9


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