anonymous
  • anonymous
How do you factor 4x^3-13x^2+3x? I started by taking the GCF (x) out, and I got 4x^2-13x+3, but after that, I'm not sure what to do.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Hero
  • Hero
If you factored out x then you have \[x(4x^2 - 13x + 3)\] Now find two numbers, \(m\) and \(n\) that add to get -13, yet multiply to get 12. \[m + n = -13 \\ m \times n = 12\]
anonymous
  • anonymous
-12 and -1 I tried that, but then when I put them in factored for, I tried checking for the answer and it didn't work. (4x-1) (4x-12)
Hero
  • Hero
Okay, I see that your factorization is not correct.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Hero
  • Hero
Here's what you're supposed to do after finding the two numbers \(x(4x^2 -13x + 3)=x(4x^2 -12x - 1x + 3)\) \(=x(4x(x - 3) -1(x - 3))\) \(=x((x - 3)(4x - 1))\) \(=x(x - 3)(4x - 1)\)
anonymous
  • anonymous
THANK YOU! I wasn't taught that method, but now that you've shown me, THANKS! If this were any other problem where I didn't have a GCF, could I still keep the -12x and -1x separate?
Hero
  • Hero
Yes. The method is called factor by grouping. If the quadratic is factorable, then you can use that method. However, you have to know what you're doing in order to use it.
anonymous
  • anonymous
Yup, I understand, thank you!
anonymous
  • anonymous
Quick question, sorry to come back to this problem again, but the -1x+3... when you factored the -1 out, are you allowed to change the sign from a + to a - inside the parentheses of the x+3?
Hero
  • Hero
Basically.... Suppose you had \[-1(x - 3)\] And you were asked to use the distributive property to expand the expression...what would be the result?
Hero
  • Hero
If you were to expand the expression using the distributive property, you would see that you arrive back at the original expression which is \(-1x + 3\) because \(-1(x - 3) = -1(x) -1(-3)= -1x + 3\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.