What is (3t^4+4t^3-32t^2-5t-20)(t+4)^{-1} ?

The solution to (3t^4+4t^3-32t^2-5t-20)(t+4)^{-1} is 3t^3-8t^2-5

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simplify (3t^4+4t^3-32t^2-5t-20)(t+4)^{-1}

simplify

(3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20) (t + 4)^{- 1}

3 t^{3} - 8 t^{2} - 5

Solution steps

(3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20) (t + 4)^{- 1}

Apply exponent rule:

a^{- 1} = \frac{1}{a}

(t + 4)^{- 1} = \frac{1}{t + 4}

= (3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20) \frac{1}{t + 4}

Apply the fraction rule:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{( 3 t ^{4} + 4 t ^{3} - 32 t ^{2} - 5 t - 20 ) \cdot 1}{t + 4}

Apply rule:

a \cdot 1 = a

(3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20) \cdot 1 = (3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20)

= \frac{3 t ^{4} + 4 t ^{3} - 32 t ^{2} - 5 t - 20}{t + 4}

Factor

3 t^{4} + 4 t^{3} - 32 t^{2} - 5 t - 20 : (t + 4) (3 t^{3} - 8 t^{2} - 5)

= \frac{( t + 4 ) ( 3 t ^{3} - 8 t ^{2} - 5 )}{t + 4}

Cancel the common factor:

t + 4

= 3 t^{3} - 8 t^{2} - 5

8 x^{7}

\frac{- 20}{5}

2^{x} \cdot 5^{x}

e^{- x} (e^{- x})

(x - 2)^{3} (x + 3)^{3}

What is (3t^4+4t^3-32t^2-5t-20)(t+4)^{-1} ?
The solution to (3t^4+4t^3-32t^2-5t-20)(t+4)^{-1} is 3t^3-8t^2-5