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Mathematics : Partial Fractions Questions

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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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Anonymous
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Anonymous
Thanks for the help.

Anonymous
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