Question

(5a³+6a⁴+3a²-9+5a)÷(a²-1)​

Answer 1

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Answer 2

SOLUTION

CORRECT QUESTION

TO DETERMINE

$\displaystyle \sf{ \frac{5 {a}^{3} + 6 {a}^{4} + 3 {a}^{2} - 9 - 5a }{ {a}^{2} - 1} }$

EVALUATION

Here the given fraction is

$\displaystyle \sf{ = \frac{5 {a}^{3} + 6 {a}^{4} + 3 {a}^{2} - 9 - 5a }{ {a}^{2} - 1} }$

Which can be rewritten as

$\displaystyle \sf{ = \frac{6 {a}^{4} +5 {a}^{3} + 3 {a}^{2} - 5a - 9}{ {a}^{2} - 1} }$

Now Numerator

$\sf{ = 6 {a}^{4} +5 {a}^{3} + 3 {a}^{2} - 5a - 9}$

For a = 1 the value of numerator is zero

∴ ( a - 1 ) is a factor of the numerator

$\sf{ 6 {a}^{4} +5 {a}^{3} + 3 {a}^{2} - 5a - 9}$

$\sf{ = 6 {a}^{4} - 6 {a}^{3} +11 {a}^{3} - 11 {a}^{2} + 14 {a}^{2} - 14a + 9a - 9}$

$\sf{ = 6 {a}^{3}(a - 1) +11 {a}^{2} (a - 1) + 14a(a -1) + 9(a - 1)}$

$\sf{ =(a - 1) (6 {a}^{3} +11 {a}^{2} + 14a+ 9)}$

$\sf{ =(a - 1) (6 {a}^{3} + 6 {a}^{2} +5 {a}^{2} + 5a + 9a+ 9)}$

$\sf{ =(a - 1) \bigg[6 {a}^{2} (a + 1) +5a(a + 1) + 9(a+1 ) \bigg] }$

$\sf{ =(a - 1) (a + 1) (6 {a}^{2} +5a + 9 ) }$

$\sf{ =( {a}^{2} - 1) (6 {a}^{2} +5a + 9 ) }$

Hence

$\displaystyle \sf{ \frac{5 {a}^{3} + 6 {a}^{4} + 3 {a}^{2} - 9 - 5a }{ {a}^{2} - 1} }$

$\displaystyle \sf{ = \frac{6 {a}^{4} +5 {a}^{3} + 3 {a}^{2} - 5a - 9}{ {a}^{2} - 1} }$

$\displaystyle \sf{ = \frac{({a}^{2} - 1) (6 {a}^{2} +5a + 9 )}{ {a}^{2} - 1} }$

$\displaystyle \sf{ = (6 {a}^{2} +5a + 9 ) }$

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