ax^2+bx+8=(3x+2)(x+k), find a, b and k

[momoko]

If ax[sup:10iaikbo]2[/sup:10iaikbo] + bx + 8 = (3x + 2)(x + k), find the values of a, b, and k.

This is supposed to be my practice question, and I only know that the final answer is a=3, b=14 and k=4
Thanks in advance! I'm in secondary 2 this year.
OK I just read the forum rules and I realise I need to write out the steps I did.
ax^2+bx+8=(3x+2)(x+k)
ax^2+bx+8=3x^2+3kx+2x+2k
ax^2+bx+8-(3x^2+3kx+2x+2k)=0
ax^2+bx+8-3x^2-3kx-2x-2k=0
-3x^2-2x+8=ax^2-bx+3kx+2k
(-3x+4)(x+2)=-x(ax-b)+k(3x+2)

I don't seem to be anywhere close to the solution, what am I supposed to do? Thanks

 

[Denis]

If ax[sup:6zs26l8l]2[/sup:6zs26l8l] + bx + 8 = (3x + 2)(x + k), find the values of a, b, and k.

When you got to this point:
ax^2+bx+8=3x^2+3kx+2x+2k
simply do this step:
ax^2 + bx + 8 = 3x^2 + x(3k +2) + 2k

Then you easily "see" that:
a = 3 : x^2 terms
2k = 8 : 3rd terms ; so k = 4

b = 3k + 2 : x terms
So b = 3(4) + 2 = 14

OK??

ALWAYS look for short-cuts with such problems :idea:

 

[momoko]

If ax[supk926hrn]2[/supk926hrn] + bx + 8 = (3x + 2)(x + k), find the values of a, b, and k.

When you got to this point:
ax^2+bx+8=3x^2+3kx+2x+2k
simply do this step:
ax^2 + bx + 8 = 3x^2 + x(3k +2) + 2k

Then you easily "see" that:
a = 3 : x^2 terms
2k = 8 : 3rd terms ; so k = 4

b = 3k + 2 : x terms
So b = 3(4) + 2 = 14

OK??

ALWAYS look for short-cuts with such problems :idea:

Thank you very much! I understand now