What is (4x^2-3x+8)-(2x^2+2x-5) ?

The solution to (4x^2-3x+8)-(2x^2+2x-5) is 2x^2-5x+13

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simplify (4x^2-3x+8)-(2x^2+2x-5)

simplify

(4 x^{2} - 3 x + 8) - (2 x^{2} + 2 x - 5)

2 x^{2} - 5 x + 13

Solution steps

(4 x^{2} - 3 x + 8) - (2 x^{2} + 2 x - 5)

Apply rule:

(a) = a

(4 x^{2} - 3 x + 8) = 4 x^{2} - 3 x + 8

= 4 x^{2} - 3 x + 8 - (2 x^{2} + 2 x - 5)

Apply the distributive law:

- (a + b) = - a - b

- (2 x^{2} + 2 x - 5) = - 2 x^{2} - 2 x + 5

= 4 x^{2} - 3 x + 8 - 2 x^{2} - 2 x + 5

Group like terms

= 4 x^{2} - 2 x^{2} - 3 x - 2 x + 8 + 5

Add similar elements:

4 x^{2} - 2 x^{2} = 2 x^{2}

= 2 x^{2} - 3 x - 2 x + 8 + 5

Add similar elements:

- 3 x - 2 x = - 5 x

= 2 x^{2} - 5 x + 8 + 5

Add the numbers:

8 + 5 = 13

= 2 x^{2} - 5 x + 13

2 - (- \frac{2}{3})

(5 x^{2} + 4)^{2}

\frac{40}{11}

x^{2} y (- 2 x^{5} y^{5})

\frac{5}{9} + \frac{7}{9}

What is (4x^2-3x+8)-(2x^2+2x-5) ?
The solution to (4x^2-3x+8)-(2x^2+2x-5) is 2x^2-5x+13