anonymous
  • anonymous
One factor of 4x^3+15x^2-31x-30 is x-2. Find the other factors
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.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
You could use long division or synthetic division. Your choice.
anonymous
  • anonymous
how
anonymous
  • anonymous
with long division

Looking for something else?

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

More answers

anonymous
  • anonymous
Well to start with you know that \[4x^2 \times (x-2) = 4x^3-8x^2\] so just like you would do long division with numbers you can also do it with variables like this. so \[ 4x^3-15x^2 -\bigg[4x^3-8x^2 \bigg]=-7x^2\] you drop the -31x like long division to get \[-7x^2 -31x\] and then divide this by (x-2) again \[-7x\times (x-2) = -7x^2 +14x\] and so on
anonymous
  • anonymous
sorry the \[-15x^2 \] should be positive 15
anonymous
  • anonymous
so \[4x^3+15x^2 -\bigg[4x^3-8x^2 \bigg]=23x^2\] \[23x\times (x-2) = 23x^2-46x\] \[23x^2-31x -\bigg[23x^2-46x\bigg]=+15x\] \[15\times(x-2)=15x-30\] \[+15x-30 -\bigg[15x-30 \bigg]=0\] so \[4x^3+15x^2-31x-30 \to (x-2)(4x^2+23x+15)\] and you can factor the rest... hopefully
anonymous
  • anonymous
Explaining how to do long division or synthetic division is a real pain on open study, I would suggest you take the time outside of this to study it from a book with examples. Or go online and search for a youtube video. It isn't hard.
anonymous
  • anonymous
thank you

Looking for something else?

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