Solve the given 125+4X-9x2 State the type of partial fraction (x+1)(x-1)(x+2) a) Linear repeated factor b) Linearnon repeated factor c) Irreducible repeated quadratic factor d) Irreducible non repeated quadratic factor
Answers
Answer:
The factors of the denominator are \displaystyle x,\left({x}^{2}+1\right)x,(x
2
+1), and \displaystyle {\left({x}^{2}+1\right)}^{2}(x
2
+1)
2
. Recall that, when a factor in the denominator is a quadratic that includes at least two terms, the numerator must be of the linear form \displaystyle Ax+BAx+B. So, let’s begin the decomposition.
\displaystyle \frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{A}{x}+\frac{Bx+C}{\left({x}^{2}+1\right)}+\frac{Dx+E}{{\left({x}^{2}+1\right)}^{2}}
x(x
2
+1)
2
x
4
+x
3
+x
2
−x+1
=
x
A
+
(x
2
+1)
Bx+C
+
(x
2
+1)
2
Dx+E
We eliminate the denominators by multiplying each term by \displaystyle x{\left({x}^{2}+1\right)}^{2}x(x
2
+1)
2
. Thus,
\displaystyle {x}^{4}+{x}^{3}+{x}^{2}-x+1=A{\left({x}^{2}+1\right)}^{2}+\left(Bx+C\right)\left(x\right)\left({x}^{2}+1\right)+\left(Dx+E\right)\left(x\right)x
4
+x
3
+x
2
−x+1=A(x
2
+1)
2
+(Bx+C)(x)(x
2
+1)+(Dx+E)(x)
Expand the right side.
x
4
+
x
3
+
x
2
−
x
+
1
=
A
(
x
4
+
2
x
2
+
1
)
+
B
x
4
+
B
x
2
+
C
x
3
+
C
x
+
D
x
2
+
E
x
=
A
x
4
+
2
A
x
2
+
A
+
B
x
4
+
B
x
2
+
C
x
3
+
C
x
+
D
x
2
+
E
x
x4+x3+x2−x+1=A(x4+2x2+1)+Bx4+Bx2+Cx3+Cx+Dx2+Ex =Ax4+2Ax2+A+Bx4+Bx2+Cx3+Cx+Dx2+Ex
Now we will collect like terms.
\displaystyle {x}^{4}+{x}^{3}+{x}^{2}-x+1=\left(A+B\right){x}^{4}+\left(C\right){x}^{3}+\left(2A+B+D\right){x}^{2}+\left(C+E\right)x+Ax
4
+x
3
+x
2
−x+1=(A+B)x
4
+(C)x
3
+(2A+B+D)x
2
+(C+E)x+A
Set up the system of equations matching corresponding coefficients on each side of the equal sign.
A
+
B
=
1
C
=
1
2
A
+
B
+
D
=
1
C
+
E
=
−
1
A
=
1
A+B=1 C=12A+B+D=1 C+E=−1 A=1
We can use substitution from this point. Substitute \displaystyle A=1A=1 into the first equation.
1
+
B
=
1
B
=
0
1+B=1 B=0
Substitute \displaystyle A=1A=1 and \displaystyle B=0B=0 into the third equation.
2
(
1
)
+
0
+
D
=
1
D
=
−
1
2(1)+0+D=1 D=−1
Substitute \displaystyle C=1C=1 into the fourth equation.
1
+
E
=
−
1
E
=
−
2
1+E=−1 E=−2
Now we have solved for all of the unknowns on the right side of the equal sign. We have \displaystyle A=1A=1, \displaystyle B=0B=0, \displaystyle C=1C=1, \displaystyle D=-1D=−1, and \displaystyle E=-2E=−2. We can write the decomposition as follows:
\displaystyle \frac{{x}^{4}+{x}^{3}+{x}^{2}-x+1}{x{\left({x}^{2}+1\right)}^{2}}=\frac{1}{x}+\frac{1}{\left({x}^{2}+1\right)}-\frac{x+2}{{\left({x}^{2}+1\right)}^{2}}
x(x
2
+1)
2
x
4
+x
3
+x
2
−x+1
=
x
1
+
(x
2
+1)
1
−
(x
2
+1)
2
x+2
TRY IT 4
Find the partial fraction decomposition of the expression with a repeated irreducible quadratic factor.
\displaystyle \frac{{x}^{3}-4{x}^{2}+9x - 5}{{\left({x}^{2}-2x+3\right)}^{2}}
(x
2
−2x+3)
2
x
3
−4x
2
+9x−5
Solution
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