How do you factor completely #x^2-12x+35#?

3 Answers
Mar 17, 2018

Answer:

#(x-5)(x-7)=>x=5, x=7#

Explanation:

We need to find two numbers that add up to #-12# and those same two numbers have a product of #35#. After some thinking and trial and error, we come up with

#-5+ -7=-12# & #-5*-7=35#

Therefore #-5# and #-7# will be our two factors. We have:

#(x-5)(x-7)=0#

as our factored form, and to find the zeros, take the opposite sign to get:

#x=5# and #x=7#

Mar 17, 2018

Answer:

#(x-7)(x-5)#

Explanation:

#x^2 - 12x + 35#

So what my math teacher used to teach me is that the place where the #35# is, that's where numbers multiply and their product is #35#. Where the #-12# is, that's where numbers add up and its result is #-12#.

However, the numbers have to be the same, so what do we do?

We first list all the factors of #35#:

  • #1 * 35#
  • #5 * 7#
  • #7 * 5#
  • #35 * 1#

And then the negatives (wink-wink it's going to be negatives...)

  • #-1 * -35#
  • #-5 * -7#
  • #-7 * -5#
  • #-35 * -1#

And then the one negative one positive... and so on. However, did you noticed that the middle number is negative?

A positive times positive can only have a positive result. That is eliminated. If it is a negative time a positive, then it can only be:

  • #-35 + 1 = -34#
  • #-1 + 35 = 34#
  • #-7 + 5 = -2#
  • #-5 + 7 = 2#

That are the only possible answers. But none of them are #-12#.
So the only way is negative times a negative. And the only combination that results in a #-12# is: #-7# and #-5#. Added together is #-12#. Multiplied together is #35#.

Voila, you have your answer. And of course, you put it with the #x# also; don't forget that!!

Answer: #(x-7)(x-5)#

Double check:

  • #x * x#
  • #-7x -5x#
  • #-7 * -5#

so

#x^2 - 12x + 35#

Hope this helps!! :)

Mar 17, 2018

Answer:

Find the 2 factors of 35 that also add to -12

Explanation:

Because our equation is already in the form #ax^2+bx+c#, we are ready to begin factoring.

Begin by finding the factors that multiply to 35 and add to -12.

Personally, if I don't see the factors after a few seconds I will begin listing ALL the factors of the "c" term and begin fiddling with the signs of the individual factors until I find something that works.

Factors: #(1, 35), (5,7)# are our only factors. We know that 5 and 7 multiply to 35 and add to 12. But, if we take the negative versions of both of these factors it still multiplies to 35 and adds to -12. These are our factors.

We then take those factors and rewrite our equation in the following form

#(x-5)*(x-7)#

These are the factors that you will be looking for. To check our work, simply distribute to undo the factoring process.