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

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Answer 1 · 2016-11-09T22:41:28

$- 11 x + 10$ $-11x + 10$

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

$x^{2} - 6 x + 1 + 2 x^{2} - 5 x + 5 \Rightarrow$ $x^2 - 6x + 1 + 2x^2 - 5x + 5 =>$

$x^{2} + 2 x^{2} - 6 x - 5 x + 1 + 5 \Rightarrow$ $x^2 + 2x^2 - 6x - 5x + 1 + 5 =>$

$3 x^{2} - 11 x + 6 \Rightarrow$ $3x^2 - 11x + 6 =>$

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

$3 x^{2} - 11 x + 6 - (3 x^{2} - 4) \Rightarrow$ $3x^2 - 11x + 6 - (3x^2 - 4) =>$

$3 x^{2} - 11 x + 6 - 3 x^{2} + 4) \Rightarrow$ $3x^2 - 11x + 6 - 3x^2 + 4) =>$

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

$- 11 x + 10$ $-11x + 10$

Answer 2 · 2016-11-09T22:50:15

$- 11 x + 10$ $" "-11x+10$

Using the old fashioned approach of lining things up.

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

$2 x^{2} - 6 x + 1$ $color(white)(2)x^2-6x+1$
$ul(2x^2-5x+5)$
$color(red)(3x^2-11x+6)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now subtract $color(green)(3x^3-4)$

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

$color(red)(3x^2-11x+6)$
$color(green)(ul(3x^2+color(white)(1)0x-4)) larr" Subtract"$
$0x^2-11x+10$

Giving: $" "-11x+10$

How do you subtract 3x^2-43x2−43x^2-4 from the sum of x^2-6x+1x2−6x+1x^2-6x+1 and 2x^2-5x+52x2−5x+52x^2-5x+5?