jippie771h

2021-12-06

Factor $2{x}^{2}-8x-10$ completely.

Mary Darby

Step 1
Given,
$2{x}^{2}-8x-10$
To split the middle term:
Multiplying the coefficient of ${x}^{2}$ and the constant term $=-20$
The coefficient of $=-8$
We want the product of two numbers to be -20 and sum of the same two numbers to be -8
Cleartly, the numbers are -10 and 2
Step 2
Therefore, we split the middle term accordingly, we have
$2{x}^{2}-8x-10$
$=2{x}^{2}-10+2x-10$
Taking 2x common from ${1}^{st}$ two terms and 2 common from last two terms, we have
$=2x\left(x-5\right)+2\left(x-5\right)$
Again, taking $x-5$ common from the complete expression, we have
$=\left(x-5\right)\left(2x+2\right)$
$=2\left(x+1\right)\left(x-5\right)$
Therefore, $2{x}^{2}-8x-10=2\left(x+1\right)\left(x-5\right)$

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