Question

(2x-7)-(4x^2+6x-3)+(x^2+4)

Answer 1

Answer:

$-3x^2-4x$

Step-by-step explanation:

Simplify:

$\left(2x-7\right)-\left(4x^2+6x-3\right)+\left(x^2+4\right)$

$=2x-7-\left(4x^2+6x-3\right)+x^2+4$

Group same factor:

$=-\left(4x^2+6x-3\right)+x^2+2x-7+4$

Add :

$=-\left(4x^2+6x-3\right)+x^2+2x-3$

Apply the distributive law: -(a+b) = -a - b

$-\left(4x^2+6x-3\right)=-4x^2-6x+3$

$=-4x^2-6x+3+x^2+2x-3$

Group same factor:

$=-4x^2+x^2-6x+2x+3-3$

Subtract:

$=-4x^2+x^2-6x+2x$

Add:

$=-3x^2-6x+2x$

Group same factor:

$=-3x^2-4x$

Hence answer = $-3x^2-4x$

~lenvy~