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IrinaVladis [17]
3 years ago
15

Simplify completely x^2+4x-45 / x^2+10x+9

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0

Answer: x - 5/x + 1

Step-by-step explanation:

This algebraic fraction

The task to be performed here is factorisation and simplification. Now going by the question,

x² + 4x - 45/x² + 10x + 9, the factorisation of

x² + 4x - 45 = x² + 9x - 5x - 45

= x(x + 9 ) - 5(x + 9 )

= ( x + 9 )(x - 5 ), don't forget this is the algebraic fraction's Numerator

The second part

x² + 10x + 9 = x² + x + 9x + 9

= x(x + 1) + 9( x + 1 )

= ( x + 9 )( x + 1 ), this is the algebraic denominator.

Now place the second expression which is the denominator under the first expression which is the numerator.

( x + 9 )( x - 5 )/( x + 9 )( x + 1 ).

You can see that, ( x + 9 )/( x + 9 ) divide each other , therefore therr then cancelled and left with

x - 5/x + 1

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          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

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