What is expand (2x+3)(4x+4Δx-5) ?

The solution to expand (2x+3)(4x+4Δx-5) is 8x^2+8Δx^2+2x+12Δx-15

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expand (2x+3)(4x+4Δx-5)

expand

(2 x + 3) (4 x + 4Δ x - 5)

8 x^{2} + 8Δ x^{2} + 2 x + 12Δ x - 15

Solution steps

(2 x + 3) (4 x + 4Δ x - 5)

Distribute parentheses

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

Apply minus-plus rules

+ (- a) = - a

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

Simplify

2 \cdot 4 xx + 2 \cdot 4Δ xx - 2 \cdot 5 x + 3 \cdot 4 x + 3 \cdot 4Δ x - 3 \cdot 5 : 8 x^{2} + 8Δ x^{2} + 2 x + 12Δ x - 15

= 8 x^{2} + 8Δ x^{2} + 2 x + 12Δ x - 15

\frac{5 ( s + 2 )}{s ^{2} ( s + 1 ) ( s + 5 )}

\frac{1}{( 3 - z ) ^{4}}

π^{e^{0}} - arccos (0)

(2 - j^{2})^{8}

w + 14 w - 6 w

What is expand (2x+3)(4x+4Δx-5) ?
The solution to expand (2x+3)(4x+4Δx-5) is 8x^2+8Δx^2+2x+12Δx-15