Factor.
-$$6 \mathrm { x } ^ { 2 } + 11 \mathrm { xt } + 4 \mathrm { t } ^ { 2 }$$
A) $( 2 x - t ) ( 3 x - 4 t )$
B) $( 2 x + t ) ( 3 x + 4 t )$
C) $( 6 x + t ) ( x + 4 t )$
D) Prime

Question

Quizplus · Accepted Answer

The correct factorization of $6x^2 + 11xt + 4t^2$ is found by looking for two numbers that multiply to $6 \times 4 = 24$ and add to 11 (the coefficient of the middle term). These numbers are 8 and 3. Rewriting the middle term $11xt$ as $8xt + 3xt$ and factoring by grouping gives us $(2x + t)(3x + 4t)$.

Factor - 6x2+11xt+4t26 \mathrm { x } ^ { 2 } + 11 \mathrm { xt } + 4 \mathrm { t } ^ { 2 }6x2+11xt+4t2

Factor
- $6 \mathrm { x } ^ { 2 } + 11 \mathrm { xt } + 4 \mathrm { t } ^ { 2 }$