What is 2*(b^2-b+7)-(b^2+b+8) ?

The solution to 2*(b^2-b+7)-(b^2+b+8) is b^2-3b+6

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simplify 2*(b^2-b+7)-(b^2+b+8)

simplify

2 \cdot (b^{2} - b + 7) - (b^{2} + b + 8)

b^{2} - 3 b + 6

Solution steps

2 (b^{2} - b + 7) - (b^{2} + b + 8)

Apply the distributive law:

- (a + b) = - a - b

- (b^{2} + b + 8) = - b^{2} - b - 8

= 2 (b^{2} - b + 7) - b^{2} - b - 8

Expand

2 (b^{2} - b + 7) : 2 b^{2} - 2 b + 14

= 2 b^{2} - 2 b + 14 - b^{2} - b - 8

Group like terms

= 2 b^{2} - b^{2} - 2 b - b + 14 - 8

Add similar elements:

2 b^{2} - b^{2} = b^{2}

= b^{2} - 2 b - b + 14 - 8

Add similar elements:

- 2 b - b = - 3 b

= b^{2} - 3 b + 14 - 8

Subtract the numbers:

14 - 8 = 6

= b^{2} - 3 b + 6

- 2 [7 a - (- 7 a + 5 b - 1 + 6 (a - 8 b + 9)) - (4 b - 8)]

52 \cdot 96

(1 - cos^{2} (x))^{2}

\frac{1 + i}{1 - \frac{1 + i}{1 - \frac{1 - i}{1 + i}}}

tan (x) cot (x) - cos^{2} (x)

What is 2*(b^2-b+7)-(b^2+b+8) ?
The solution to 2*(b^2-b+7)-(b^2+b+8) is b^2-3b+6