What is expand (x^2-2)^4 ?

The solution to expand (x^2-2)^4 is x^8-8x^6+24x^4-32x^2+16

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expand (x^2-2)^4

expand

(x^{2} - 2)^{4}

x^{8} - 8 x^{6} + 24 x^{4} - 32 x^{2} + 16

Solution steps

(x^{2} - 2)^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = x^{2}, b = - 2

= i = 0 \sum 4 (i 4) (x^{2})^{(4 - i)} (- 2)^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (x^{2})^{4} (- 2)^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (x^{2})^{3} (- 2)^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (x^{2})^{2} (- 2)^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (x^{2})^{1} (- 2)^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (x^{2})^{0} (- 2)^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (x^{2})^{4} (- 2)^{0} : x^{8}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (x^{2})^{3} (- 2)^{1} : - 8 x^{6}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (x^{2})^{2} (- 2)^{2} : 24 x^{4}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (x^{2})^{1} (- 2)^{3} : - 32 x^{2}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (x^{2})^{0} (- 2)^{4} : 16

= x^{8} - 8 x^{6} + 24 x^{4} - 32 x^{2} + 16

- 32 \div 4

(- 8) (- 9)

2 + 42

\frac{6}{64}

2 y^{2} - 5 y + 3

What is expand (x^2-2)^4 ?
The solution to expand (x^2-2)^4 is x^8-8x^6+24x^4-32x^2+16