Factorize  3×2+7x-6 by splitting the middle term.

# Factorize  $3{x}^{2}+7x-6$ by splitting the middle term.

1. A
$\left(3x+2\right)\left(x+3\right)$
2. B
$\left(3x-2\right)\left(x+3\right)$
3. C
$\left(3x-2\right)\left(x-3\right)$
4. D
$\left(3x+2\right)\left(x-3\right)$

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### Solution:

The given polynomial is $3{x}^{2}+7x-6.$
Comparing the given polynomial with $a{x}^{2}+\mathit{bx}+c$ we get,
$a=3,b=7,c=-6$ .
Here $\mathit{ac}=\left(3\right)\left(-6\right)=-18$  , so we try to split $b=7$ into two parts whose sum is 7 and product is -18.
Therefore, possible factors are and 2, -2 and 9.
Clearly, pair -2 and 9 gives $-2+9=7=b$
$=3{x}^{2}+\left(-2+9\right)x-6$
$=3{x}^{2}-2x+9x-6$
$=3x\left(x+3\right)-2\left(x+3\right)$
$=\left(3x-2\right)\left(x+3\right)$ Thus, the factors of $3{x}^{2}+7x-6$ , by using splitting the middle term are.
Therefore, option 2 is correct.

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