Factoring 6x^2-15x-21: got to 3(x^2 - 5x - 7), and got stuck

[spicytakoyum]

Hello,

I started working on this factoring problem:

6x^2-15x-21

then started factoring to

3(x^2-5x-7)

and now I'm stuck- did I even start right?

Thank you,
Mabel

 

[Deleted member 4993]

Re: Factoring...

I started working on this factoring problem:

6x^2-15x-21

then started factoring to

3(2x^2-5x-7).............Now you can factorize further by realizing

-5 = -7 + 2

 

[jwpaine]

Further on Subhotosh's post:

You can further factor this by using the grouping method:

\(\displaystyle \L 2x^{2}\,-5x\,- \,7\,\,\,=\,\,\, 2x^{2}\,-7x\,+\,2x\,-7\)

And -7 and 2 are both divisible by (a)(c) = -14

You can now group \(\displaystyle \L 2x^{2}\,-7x\,+\,2x\,-7\) into a (binomial) + (binomial), Continue to factor out terms from each binomial and then you can further factor so that you have a (binomial)(binomial), and thus 3(binomial)(binomial) = \(\displaystyle \L 2x^{2}\,-7x\,+\,2x\,-7\) in factored form.

John.