What is (3x^{n-1})(x^{n+1})(4x^{2-n}) ?

The solution to (3x^{n-1})(x^{n+1})(4x^{2-n}) is 12x^{n+2}

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

simplify

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

12 x^{n + 2}

Solution steps

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

Apply rule:

(a) = a

(x^{n + 1}) = x^{n + 1}

= 3 x^{n - 1} x^{n + 1} \cdot 4 x^{2 - n}

Multiply the numbers:

3 \cdot 4 = 12

= 12 x^{n - 1} x^{n + 1} x^{2 - n}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

x^{n - 1} x^{n + 1} x^{2 - n} = x^{n - 1 + n + 1 + 2 - n}

= 12 x^{n - 1 + n + 1 + 2 - n}

Simplify

n - 1 + n + 1 + 2 - n : n + 2

= 12 x^{n + 2}

3 x^{4} - 6 x^{3} y

x^{2} - (x + 1) (x - 2)

\frac{2 x}{3} (12 x + 18)

(4 a^{2} - a + 3)^{2}

- 12 \cdot 5

What is (3x^{n-1})(x^{n+1})(4x^{2-n}) ?
The solution to (3x^{n-1})(x^{n+1})(4x^{2-n}) is 12x^{n+2}