Factor using the ac-method.
-$8 x^{2}+14 x t+3 t^{2}$
A) $(8 \mathrm{x}+\mathrm{t})(\mathrm{x}+3 \mathrm{t})$
B) $(4 \mathrm{x}+\mathrm{t})(2 \mathrm{x}+3 \mathrm{t})$
C) $(4 x-t)(2 x-3 t)$
D) prime

Question

Quizplus · Accepted Answer

The ac-method involves multiplying the coefficient of $x^2$ (which is 8) by the constant term (which is $3t^2$), giving $24t^2$. We then find two numbers that multiply to $24t^2$ and add to the middle coefficient $14t$. These numbers are $6t$ and $8t$. We then rewrite the middle term $14xt$ as $6xt + 8xt$ and factor by grouping, which gives us $(4x+t)(2x+3t)$.

Factor Using the Ac-Method 8x2+14xt+3t28 x^{2}+14 x t+3 t^{2}8x2+14xt+3t2 A) (8x+t)(x+3t)(8 \mathrm{x}+\mathrm{t})(\mathrm{x}+3 \mathrm{t})(8x+t)(x+3t)

Factor Using the Ac-Method $8 x^{2}+14 x t+3 t^{2}$ A) $(8 \mathrm{x}+\mathrm{t})(\mathrm{x}+3 \mathrm{t})$