# How do you subtract 3x^2-4 from the sum of x^2-6x+1 and 2x^2-5x+5?

Nov 9, 2016

$- 11 x + 10$

#### Explanation:

The sum of ${x}^{2} - 6 x + 1$ and $2 {x}^{2} - 5 x + 5$ is solved as:

${x}^{2} - 6 x + 1 + 2 {x}^{2} - 5 x + 5 \implies$

${x}^{2} + 2 {x}^{2} - 6 x - 5 x + 1 + 5 \implies$

$3 {x}^{2} - 11 x + 6 \implies$

Then to subtract $3 {x}^{2} - 4$ from this is solved as:

$3 {x}^{2} - 11 x + 6 - \left(3 {x}^{2} - 4\right) \implies$

3x^2 - 11x + 6 - 3x^2 + 4) =>

$3 {x}^{2} - 3 {x}^{2} - 11 x + 6 + 4$

$- 11 x + 10$

Nov 9, 2016

$\text{ } - 11 x + 10$

#### Explanation:

Using the old fashioned approach of lining things up.

The sum of $\left({x}^{2} - 6 x + 1\right) + \left(2 {x}^{2} - 5 x + 5\right)$

$\textcolor{w h i t e}{2} {x}^{2} - 6 x + 1$
$\underline{2 {x}^{2} - 5 x + 5}$
$\textcolor{red}{3 {x}^{2} - 11 x + 6}$

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Now subtract $\textcolor{g r e e n}{3 {x}^{3} - 4}$

I am using the place holder of no values of $x$ as $0 x$ to make things line up properly.

$\textcolor{red}{3 {x}^{2} - 11 x + 6}$
$\textcolor{g r e e n}{\underline{3 {x}^{2} + \textcolor{w h i t e}{1} 0 x - 4}} \leftarrow \text{ Subtract}$
$0 {x}^{2} - 11 x + 10$

Giving: $\text{ } - 11 x + 10$