What is expand 4(x+Δx)^2-3(x+Δx) ?

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

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expand 4(x+Δx)^2-3(x+Δx)

expand

4 (x + Δ x)^{2} - 3 (x + Δ x)

4 x^{2} + 8Δ x^{2} + 4 Δ^{2} x^{2} - 3 x - 3Δ x

Solution steps

4 (x + Δ x)^{2} - 3 (x + Δ x)

Expand

- 3 (x + Δ x) : - 3 x - 3Δ x

= 4 (x + Δ x)^{2} - 3 x - 3Δ x

Simplify

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

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

((3 x^{2} - 9)^{3}) x d x

(- 4) (2 x - 3 y = 9)

ln (z) (z - 1)

b 3 + b 2

\frac{( n + 5 ) ^{2}}{( 3 n )}

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