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Business and Management

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Q -1 x+25/ x(x+5)
ANS_1 x+25/x(x+5) = A/x + B/(x+5) ……………….1
Multiply by x(x+5) on both sides, we get
x+25 = A(x+5) + Bx ……………2
Put x+5 = 0 or x = -5 in equation 2, then
-5+25 = A(-5+5) – 5B
20 = -5B
B = -4
Now put x = 0 in equation 2, then
25 = 5A
A = 5
Now putting the values of A and B in equation 1, we get
HENCE…. x+25/x(x+5) = 5/x -4/(x+5)
Q -2 x^5-3x^4+3x^3-4x^2+10x+13 / (x-2)^2*(x^2+2)
ANS_2 Multiply 13/(x-2)^2 by (x^2+2)
We find roots(zeroes) of F(x) = (x^2+2)
Polynomial roots calculator is a method which aim at finding the values
of x for which F(x) = 0
rational root test is a test which would only find rational roots that is number x
which be written as quotient of two integers. Rational root theorem states that if
a polynomial (zeroes) for a rational number P/Q then “P” is a factor of trailing
constant and “Q” is a factor of the leading co-efficients.

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With this case, the leading co-efficient is “1” and the trailing constant is “2”.
Then the factors will be :
of the leading coefficient : “1”
of the trailing constant : “1” ,”2”
Let us test ....
With the help of we find no rational roots.
(((((x^5)-(3•(x^4)))+(3•(x^3)))-(4•(x^2)))+10x) + 13*(x^2+2)/(x-2)^2
Now simplify x^5-3x^4+3x^3-4x^2+10x + 13*(x^2+2)/(x-2)^2
Now adding a fraction to a whole we rewrite the whole as a fraction using (x-2)^2 as
the denominator :
x^5- 3x^4+3x^3-4x^2+10x = x^5-3x^4+3x^3-4x^2+10x / 1 = (x^5-3x^4+3x^3-4x^2+10x)*(x-2)^2 /
(x-2)^2
The equivalent fraction: Fraction thus generated looks different but has same value as
a whole. The common denominator: Equivalent fraction and the other, involved in the
calculation share the denominator as same.
Now we pull out like factors then:
x^5-3x^4+3x^3-4x^2+10x = x*(x^4-3x^3+3x^2-4x+10)
Now we Find the roots (zeroes) of
F(x)= x^4-3x^3+3x^2-4x+10
With this case, leading co-efficient is “1” and trailing constant is “10”.
The factor will be :
Of the leading coefficient : “1”
Of the trailing constant : “1” , “2” , “5” , “10”
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Now let us test ....
The polynomial root calculator found no roots so.
Now adding two equivalent fraction . we add two equivalent fractions which are now
have common denominators. We combine the numerators and we put the sum or
difference over common denominators then, if possible, reduce to lowest terms:
x*( x^4-3x^3+3x^2-4x+10)*(x-2)^2 + 13 * (x^2 +2) x^7-7x^6+19x^5-28x^4+38x^3-43x^2+40x+26
1*(x-2)^2 1*(x^2-4x+4)
HENCE final result x^7-7x^6+19x^5-28x^4+38x^3-43x^2+40x+26
1*(x^2-4x+4)
Q -3 3x /9x^2+30x+25
ANS_3 3x/9x^2+30x+25 = 3x/(3x+5)^2
3x/(3x+5)^2 = A/(3x+5) + Bx+C/(3x+5)^2 …………….1
Multiply by (3x+5)^2 on both sides, we get
“P”
-1
-2
-5
-10
1
2
5
10

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Unformatted Attachment Preview

Q -1 x+25/ x(x+5)ANS_1 x+25/x(x+5) = A/x + B/(x+5) ??????.1Multiply by x(x+5) on both sides, we getx+25 = A(x+5) + Bx ?????2Put x+5 = 0 or x = -5 in equation 2, then-5+25 = A(-5+5) - 5B20 = -5BB = -4Now put x = 0 in equation 2, then25 = 5AA = 5Now putting the values of A and B in equation 1, we getHENCE?. x+25/x(x+5) = 5/x -4/(x+5)Q -2 x^5-3x^4+3x^3-4x^2+10x+13 / (x-2)^2*(x^2+2)ANS_2 Multiply 13/(x-2)^2 by (x^2+2)We find roots(zeroes) of F(x) = (x^2+2)Polynomial roots calculator is a method which aim at finding the valuesof x for which F(x) = 0rational root test is a test which would only find rational roots that is number x which be written as quotient of two integers. Rational root theorem states that if a polynomial (zeroes) for a rational numberP/Qthen"P"is a factor of trailing constant and"Q"is a factor of the leading co-efficients.With this case, the leading co-efficient is"1"and the trailing constant is"2".Then the factors will be :of the leading coefficient :"1"of the trailing constant :"1" ,"2"Let us test ....With the help of we find no rational roots. (((((x^5)-(3-(x^4)))+(3-(x^3)))-(4-(x^2)))+10x) + 13*(x^2+2)/(x-2)^2Now simplify x^5-3x^4+3x^3-4x^2+10x + 13*(x^2+2)/(x-2)^2Now adding a fraction to a whole we rewrite the whole as a fraction using(x-2)^2as the denominator :x^5-3x^4+3x^3-4x^2+10x = x^5-3x^4+3x^3-4x^2+10x / 1 = (x^5-3x^4+3x^3-4x^2+10x) ...
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