At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
We can first factor all the terms by 2:
6x^2 - 10x - 4=2(3x^2-5x-2)
Then we solve 3x^2-5x-2=0.
Discriminant : 25+4*3*2=49=7^2 hence the roots are (5-7)/6 and (5+7)/6 which are -1/3 and 2
Hence 3x^2-5x-2=(x-2)(3x+1)
Thus 6x^2-10x-4=2(x-2)(3x+1)
6x^2 - 10x - 4=2(3x^2-5x-2)
Then we solve 3x^2-5x-2=0.
Discriminant : 25+4*3*2=49=7^2 hence the roots are (5-7)/6 and (5+7)/6 which are -1/3 and 2
Hence 3x^2-5x-2=(x-2)(3x+1)
Thus 6x^2-10x-4=2(x-2)(3x+1)
[tex]6 x^{2} -10x-4[/tex] =
= [tex]6 x^{2} -12x+2x-4[/tex] =
= 6x(x-2)+2(x-2) =
= (x-2)(6x+2) =
= 2(x-2)(3x+1)
or
Let's solve the equation:
[tex]6 x^{2} -10x-4[/tex] = 0
Discriminant:
Δ = [tex] 10^{2} -4*6*(-4)[/tex] = 196 = [tex] 14^{2} [/tex]
The roots are:
x1 = [tex] \frac{10+14}{2*6} [/tex] = [tex] \frac{24}{12} [/tex] = 2
x2 = [tex] \frac{10-14}{2*6} [/tex] = [tex] \frac{-4}{12} [/tex] = [tex] \frac{-1}{3} [/tex]
so:
[tex]6 x^{2} -10x-4[/tex] = 6(x-2)(x+[tex] \frac{1}{3} [/tex]) = (x-2)(6x+2) = 2(x-2)(3x+1)
= [tex]6 x^{2} -12x+2x-4[/tex] =
= 6x(x-2)+2(x-2) =
= (x-2)(6x+2) =
= 2(x-2)(3x+1)
or
Let's solve the equation:
[tex]6 x^{2} -10x-4[/tex] = 0
Discriminant:
Δ = [tex] 10^{2} -4*6*(-4)[/tex] = 196 = [tex] 14^{2} [/tex]
The roots are:
x1 = [tex] \frac{10+14}{2*6} [/tex] = [tex] \frac{24}{12} [/tex] = 2
x2 = [tex] \frac{10-14}{2*6} [/tex] = [tex] \frac{-4}{12} [/tex] = [tex] \frac{-1}{3} [/tex]
so:
[tex]6 x^{2} -10x-4[/tex] = 6(x-2)(x+[tex] \frac{1}{3} [/tex]) = (x-2)(6x+2) = 2(x-2)(3x+1)
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.