What is expand 5(x+3-5(x^2-1)=x^2+7(3-x)-1) ?

The solution to expand 5(x+3-5(x^2-1)=x^2+7(3-x)-1) is -25x^2+5x+40=x^2-35x+100

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expand 5(x+3-5(x^2-1)=x^2+7(3-x)-1)

expand

5 (x + 3 - 5 (x^{2} - 1) = x^{2} + 7 (3 - x) - 1)

- 25 x^{2} + 5 x + 40 = x^{2} - 35 x + 100

Solution steps

5 (x + 3 - 5 (x^{2} - 1) = x^{2} + 7 (3 - x) - 1)

= 5 (x + 3 - 5 = x^{2} (x^{2} - 1) + 7 (3 - x) - 1)

Distribute parentheses

= 5 x + 5 \cdot 3 + 5 (- 5 (x^{2} - 1) = x^{2}) + 5 \cdot 7 (3 - x) + 5 (- 1)

Apply minus-plus rules

+ (- a) = - a

5 x + 5 \cdot 3 - 5 \cdot 5 (x^{2} - 1) = x^{2} + 5 \cdot 7 (3 - x) - 5 \cdot 1

Simplify

5 x + 5 \cdot 3 - 5 \cdot 5 (x^{2} - 1) = x^{2} + 5 \cdot 7 (3 - x) - 5 \cdot 1 : - 25 x^{2} + 5 x + 40 = x^{2} - 35 x + 100

- 25 x^{2} + 5 x + 40 = x^{2} - 35 x + 100

m (9.81 - \frac{2 s}{t ^{2}})

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

(x^{2} + y^{2}) d x + (x^{2} - x y) d y

(3 + 5 i)^{4}

(1 - \frac{5}{p})^{3}

What is expand 5(x+3-5(x^2-1)=x^2+7(3-x)-1) ?
The solution to expand 5(x+3-5(x^2-1)=x^2+7(3-x)-1) is -25x^2+5x+40=x^2-35x+100