What is expand (2x+2Δx+3)(4x-5) ?

The solution to expand (2x+2Δx+3)(4x-5) is 8Δx^2+8x^2-10Δx+2x-15

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expand (2x+2Δx+3)(4x-5)

expand

(2 x + 2Δ x + 3) (4 x - 5)

8Δ x^{2} + 8 x^{2} - 10Δ x + 2 x - 15

Solution steps

(2 x + 2Δ x + 3) (4 x - 5)

Distribute parentheses

= 2 x \cdot 4 x + 2 x (- 5) + 2Δ x \cdot 4 x + 2Δ x (- 5) + 3 \cdot 4 x + 3 (- 5)

Apply minus-plus rules

+ (- a) = - a

= 2 \cdot 4 xx - 2 \cdot 5 x + 2 \cdot 4Δ xx - 2 \cdot 5Δ x + 3 \cdot 4 x - 3 \cdot 5

Simplify

2 \cdot 4 xx - 2 \cdot 5 x + 2 \cdot 4Δ xx - 2 \cdot 5Δ x + 3 \cdot 4 x - 3 \cdot 5 : 8Δ x^{2} + 8 x^{2} - 10Δ x + 2 x - 15

= 8Δ x^{2} + 8 x^{2} - 10Δ x + 2 x - 15

\frac{( x + 4 ) ( x + 8 )}{x + 5}

(2 + 1)^{n} (2 + 2)

\frac{( - 1 + r ^{3} ) ^{2}}{r ^{3}}

(5 x^{2} - 11 y^{7})^{25}

(3 e^{2} - 5 f)^{3}

What is expand (2x+2Δx+3)(4x-5) ?
The solution to expand (2x+2Δx+3)(4x-5) is 8Δx^2+8x^2-10Δx+2x-15