What is expand (x^3-4x^2-6x+1)(x^2-3x+1) ?

The solution to expand (x^3-4x^2-6x+1)(x^2-3x+1) is x^5-7x^4+7x^3+15x^2-9x+1

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expand (x^3-4x^2-6x+1)(x^2-3x+1)

expand

(x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1)

x^{5} - 7 x^{4} + 7 x^{3} + 15 x^{2} - 9 x + 1

Solution steps

(x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1)

Distribute parentheses

= x^{3} x^{2} + x^{3} (- 3 x) + x^{3} \cdot 1 - 4 x^{2} x^{2} - 4 x^{2} (- 3 x) - 4 x^{2} \cdot 1 - 6 x x^{2} - 6 x (- 3 x) - 6 x \cdot 1 + 1 \cdot x^{2} + 1 \cdot (- 3 x) + 1 \cdot 1

Simplify

x^{3} x^{2} + x^{3} (- 3 x) + x^{3} \cdot 1 - 4 x^{2} x^{2} - 4 x^{2} (- 3 x) - 4 x^{2} \cdot 1 - 6 x x^{2} - 6 x (- 3 x) - 6 x \cdot 1 + 1 \cdot x^{2} + 1 \cdot (- 3 x) + 1 \cdot 1 : x^{5} - 7 x^{4} + 7 x^{3} + 15 x^{2} - 9 x + 1

= x^{5} - 7 x^{4} + 7 x^{3} + 15 x^{2} - 9 x + 1

\frac{d}{d x} (- sin (2 x) + 1)

\frac{d}{d x} ((5 x - 3) (4 x - \frac{3}{x}))

\int \frac{x ^{3}}{x ^{4} - 8} d x

\int \frac{x ^{2}}{x ^{3} + 9} d x

\frac{d}{d x} ((a x^{2} + b x + c) e^{\frac{- x}{4}})

What is expand (x^3-4x^2-6x+1)(x^2-3x+1) ?
The solution to expand (x^3-4x^2-6x+1)(x^2-3x+1) is x^5-7x^4+7x^3+15x^2-9x+1