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How do you combine $(x^{2} - b x - 7) - (3 x^{2} + 2 x - 4)$ $(x^2-bx-7) - (3x^2+2x-4)$ ?

2 Answers

Jul 22, 2015

Group by like powers of $x$ $x$ and perform standard arithmetic to get:
$XXXX$ $color(white)("XXXX")$ $- 2 x^{2} - (b + 2) x - 3$ $-2x^2-(b+2)x -3$

Explanation:

$(x^{2} - b x - 7) - (3 x^{2} + 2 x - 4)$ $(x^2−bx−7)−(3x^2+2x−4)$
$XXXX$ $color(white)("XXXX")$ Regroup as:
$(x^{2} - 3 x^{2}) + (- b x - 2 x) + (- 7 - (- 4))$ $(x^2-3x^2) + (-bx -2x) + (-7 -(-4))$

$= - 2 x^{2} - (b + 2) x - 3$ $=-2x^2 -(b+2)x -3$

Meave60 · Alan P.

Jul 22, 2015

$(x^{2} - b x - 7) - (3 x^{2} + 2 x - 4) = - 2 x^{2} - b x - 2 x - 3$ $(x^2-bx-7)-(3x^2+2x-4)=-2x^2-bx-2x-3$

Explanation:

$(x^{2} - b x - 7) - (3 x^{2} + 2 x - 4)$ $(x^2-bx-7)-(3x^2+2x-4)$

Distribute the negative sign to each of the terms inside the second set of parentheses, then remove the parentheses.

$x^{2} - b x - 7 - 3 x^{2} - 2 x + 4$ $x^2-bx-7-3x^2-2x+4$

Group like terms.

$x^{2} - 3 x^{2} - b x - 2 x - 7 + 4$ $x^2-3x^2-bx-2x-7+4$

Combine.

$- 2 x^{2} - b x - 2 x - 3$ $-2x^2-bx-2x-3$

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