Factor the trinomial by grouping, if possible. 10 x 2 − 9 x − 9 10 x ^ { 2 } - 9 x - 9 10 x 2 − 9 x − 9 A) ( 3 x + 5 ) ( 3 x − 2 ) ( 3 x + 5 ) ( 3 x - 2 ) ( 3 x + 5 ) ( 3 x − 2 ) B) ( 5 x + 3 ) ( 2 x − 3 ) ( 5 x + 3 ) ( 2 x - 3 ) ( 5 x + 3 ) ( 2 x − 3 ) C) ( 5 x − 3 ) ( 2 x − 3 ) ( 5 x - 3 ) ( 2 x - 3 ) ( 5 x − 3 ) ( 2 x − 3 ) D) ( 5 x + 3 ) ( 2 x + 3 ) ( 5 x + 3 ) ( 2 x + 3 ) ( 5 x + 3 ) ( 2 x + 3 ) E) The trinomial is prime.

Question

Factor the trinomial by grouping, if possible.   10 x 2 − 9 x − 9 10 x ^ { 2 } - 9 x - 9 10 x 2 − 9 x − 9 A)    ( 3 x + 5 )  ( 3 x − 2 )  ( 3 x + 5 )  ( 3 x - 2 )  ( 3 x + 5 )  ( 3 x − 2 )  B)    ( 5 x + 3 )  ( 2 x − 3 )  ( 5 x + 3 )  ( 2 x - 3 )  ( 5 x + 3 )  ( 2 x − 3 )  C)    ( 5 x − 3 )  ( 2 x − 3 )  ( 5 x - 3 )  ( 2 x - 3 )  ( 5 x − 3 )  ( 2 x − 3 )  D)    ( 5 x + 3 )  ( 2 x + 3 )  ( 5 x + 3 )  ( 2 x + 3 )  ( 5 x + 3 )  ( 2 x + 3 )  E) The trinomial is prime.

Himakshi Gurjar · Accepted Answer

Final Answer : B, Explanation :  To factor by grouping, we need to find two numbers whose product is $10\cdot(-9)=-90$ and whose sum is $-9$. We can see that $-15$ and $6$ satisfy these conditions. So we rewrite the trinomial as $10x^2-15x+6x-9$. Then we group the first two terms and the last two terms: $(10x^2-15x)+(6x-9)$. We can factor $5x$ out of the first group and $3$ out of the second group: $5x(2x-3)+3(2x-3)$. We can see that $(2x-3)$ is a common factor, so we can write the expression as $(2x-3)(5x+3)$.

Factor the trinomial by grouping, if possible. $10 x ^ { 2 } - 9 x - 9$

A) $(3 x + 5) (3 x - 2)$
B) $(5 x + 3) (2 x - 3)$
C) $(5 x - 3) (2 x - 3)$
D) $(5 x + 3) (2 x + 3)$
E) The trinomial is prime.

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