What is (2a^4+232-8)-(3a^4-3a^2-9) ?

The solution to (2a^4+232-8)-(3a^4-3a^2-9) is -a^4+3a^2+233

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simplify (2a^4+232-8)-(3a^4-3a^2-9)

simplify

(2 a^{4} + 232 - 8) - (3 a^{4} - 3 a^{2} - 9)

- a^{4} + 3 a^{2} + 233

Solution steps

(2 a^{4} + 232 - 8) - (3 a^{4} - 3 a^{2} - 9)

Apply rule:

(a) = a

(2 a^{4} + 232 - 8) = 2 a^{4} + 232 - 8

= 2 a^{4} + 232 - 8 - (3 a^{4} - 3 a^{2} - 9)

Subtract the numbers:

232 - 8 = 224

= 2 a^{4} + 224 - (3 a^{4} - 3 a^{2} - 9)

Apply the distributive law:

- (a - b) = - a + b

- (3 a^{4} - 3 a^{2} - 9) = - 3 a^{4} + 3 a^{2} + 9

= 2 a^{4} + 224 - 3 a^{4} + 3 a^{2} + 9

Group like terms

= 2 a^{4} - 3 a^{4} + 3 a^{2} + 224 + 9

Add similar elements:

2 a^{4} - 3 a^{4} = - a^{4}

= - a^{4} + 3 a^{2} + 224 + 9

Add the numbers:

224 + 9 = 233

= - a^{4} + 3 a^{2} + 233

(\frac{3 x + y}{2})^{2} - (\frac{5 x}{4 + 2 y})^{2}

\frac{- 3}{x}

\frac{( y ^{2} - 4 y + 4 )}{( ( y - 2 ) ^{2} )} \cdot \frac{3}{( y + 4 )}

2^{a} + 1^{a}

(\frac{3}{6})^{n} - (\frac{3}{6})^{n}

What is (2a^4+232-8)-(3a^4-3a^2-9) ?
The solution to (2a^4+232-8)-(3a^4-3a^2-9) is -a^4+3a^2+233