Factor the trinomial completely. $6 x ^ { 2 } + 29 x y + 28 y ^ { 2 }$
A) (2 x $-$ 7 y )(3 x $-$ 4 y )
B) (2 x + 7 y )(3 x + 4 y )
C) (2 x + 7 y )(3 x $-$ 4 y )
D) (2 x $-$ 7 y )(3 x + 4 y )
E) not factorable

Question

Quizplus · Accepted Answer

The correct factorization of $6x^2 + 29xy + 28y^2$ is obtained by finding two numbers that multiply to $6 \times 28 = 168$ and add to 29. These numbers are 7 and 24, which correspond to the factors in choice B, (2x + 7y)(3x + 4y), when considering the original coefficients in the trinomial.

Factor the Trinomial Completely 6x2+29xy+28y26 x ^ { 2 } + 29 x y + 28 y ^ { 2 }6x2+29xy+28y2

Factor the Trinomial Completely $6 x ^ { 2 } + 29 x y + 28 y ^ { 2 }$