alg 2 factoring of 3x^2 - 14x + 8 need help please

Re: alg 2 factorin need help please

3x^2-14x+8 = (3x - ___)(x - ___)

You fill in the blanks, remembering that their product must be 8 and when you check by multiplying the two binomials together the cross product must be -14x.
 
Here's something to try:

\(\displaystyle 3x^2-14x+8\)

Multiply leading coefficient times constant:

\(\displaystyle 3\times8=24\)

Now find two factors of 24 that add up to -14. That'd be -12 and -2.
Replace the middle term with \(\displaystyle -12x-2x\)

\(\displaystyle 3x^2-12x-2x+8\)

Factor by grouping: \(\displaystyle 3x(x-4)-2(x-4)\)

\(\displaystyle (3x-2)(x-4)\)
 
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