jippie771h

2021-12-06

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

Mary Darby

Beginner2021-12-07Added 11 answers

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

$=2x(x-5)+2(x-5)$

Again, taking$x-5$ common from the complete expression, we have

$=(x-5)(2x+2)$

$=2(x+1)(x-5)$

Therefore,$2{x}^{2}-8x-10=2(x+1)(x-5)$

Given,

To split the middle term:

Multiplying the coefficient of

The coefficient of

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

Taking 2x common from

Again, taking

Therefore,