Question :
frac(2x^2+3x-1)((x-1)^2(x+2))
Simplify the expression
x3−3x+22x2+3x−1
Evaluate
((x−1)2(x+2))(2x2+3x−1)
Remove the parentheses
(x−1)2(x+2)2x2+3x−1
Solution
More Steps
Evaluate
(x−1)2(x+2)
Expand the expression
More Steps
Evaluate
(x−1)2
Use (a−b)2=a2−2ab+b2 to expand the expression
x2−2x×1+12
Calculate
x2−2x+1
(x2−2x+1)(x+2)
Apply the distributive property
x2×x+x2×2−2x×x−2x×2+1×x+1×2
Multiply the terms
More Steps
Evaluate
x2×x
Use the product rule an×am=an+m to simplify the expression
x2+1
Add the numbers
x3
x3+x2×2−2x×x−2x×2+1×x+1×2
Use the commutative property to reorder the terms
x3+2x2−2x×x−2x×2+1×x+1×2
Multiply the terms
x3+2x2−2x2−2x×2+1×x+1×2
Multiply the numbers
x3+2x2−2x2−4x+1×x+1×2
Any expression multiplied by 1 remains the same
x3+2x2−2x2−4x+x+1×2
Any expression multiplied by 1 remains the same
x3+2x2−2x2−4x+x+2
The sum of two opposites equals 0
More Steps
Evaluate
2x2−2x2
Collect like terms
(2−2)x2
Add the coefficients
0×x2
Calculate
0
x3+0−4x+x+2
Remove 0
x3−4x+x+2
Add the terms
More Steps
Evaluate
−4x+x
Collect like terms by calculating the sum or difference of their coefficients
(−4+1)x
Add the numbers
−3x
x3−3x+2
x3−3x+22x2+3x−1
Show Solution
Find the excluded values
x=1,x=−2
Evaluate
((x−1)2(x+2))(2x2+3x−1)
To find the excluded values,set the denominators equal to 0
(x−1)2(x+2)=0
Separate the equation into 2 possible cases
(x−1)2=0x+2=0
Solve the equation
More Steps
Evaluate
(x−1)2=0
The only way a power can be 0 is when the base equals 0
x−1=0
Move the constant to the right-hand side and change its sign
x=0+1
Removing 0 doesn't change the value,so remove it from the expression
x=1
x=1x+2=0
Solve the equation
More Steps
Evaluate
x+2=0
Move the constant to the right-hand side and change its sign
x=0−2
Removing 0 doesn't change the value,so remove it from the expression
x=−2
x=1x=−2
Solution
x=1,x=−2
Show Solution
Rewrite the fraction
3(x−1)24+9(x−1)17+9(x+2)1
Evaluate
((x−1)2(x+2))(2x2+3x−1)
Remove the parentheses
(x−1)2(x+2)2x2+3x−1
For each factor in the denominator,write a new fraction
(x−1)2?+x−1?+x+2?
Write the terms in the numerator
(x−1)2A+x−1B+x+2C
Set the sum of fractions equal to the original fraction
(x−1)2(x+2)2x2+3x−1=(x−1)2A+x−1B+x+2C
Multiply both sides
(x−1)2(x+2)2x2+3x−1×(x−1)2(x+2)=(x−1)2A×(x−1)2(x+2)+x−1B×(x−1)2(x+2)+x+2C×(x−1)2(x+2)
Simplify the expression
2x2+3x−1=(x+2)A+(x2+x−2)B+(x2−2x+1)C
Simplify the expression
More Steps
Evaluate
(x+2)A+(x2+x−2)B+(x2−2x+1)C
Multiply the terms
A(x+2)+(x2+x−2)B+(x2−2x+1)C
Multiply the terms
A(x+2)+B(x2+x−2)+(x2−2x+1)C
Multiply the terms
A(x+2)+B(x2+x−2)+C(x2−2x+1)
Expand the expression
Ax+2A+B(x2+x−2)+C(x2−2x+1)
Expand the expression
Ax+2A+Bx2+Bx−2B+C(x2−2x+1)
Expand the expression
Ax+2A+Bx2+Bx−2B+Cx2−2Cx+C
2x2+3x−1=Ax+2A+Bx2+Bx−2B+Cx2−2Cx+C
Group the terms
2x2+3x−1=(B+C)x2+(A+B−2C)x+2A−2B+C
Equate the coefficients
⎩⎨⎧2=B+C3=A+B−2C−1=2A−2B+C
Swap the sides
⎩⎨⎧B+C=2A+B−2C=32A−2B+C=−1
Solve the equation for B
⎩⎨⎧B=2−CA+B−2C=32A−2B+C=−1
Substitute the given value of B into the equation {A+B−2C=32A−2B+C=−1
{A+2−C−2C=32A−2(2−C)+C=−1
Simplify
{A+2−3C=32A−2(2−C)+C=−1
Simplify
{A+2−3C=32A−4+3C=−1
Solve the equation for A
More Steps
Evaluate
A+2−3C=3
Move the expression to the right-hand side and change its sign
A=3−(2−3C)
Subtract the terms
A=1+3C
{A=1+3C2A−4+3C=−1
Substitute the given value of A into the equation 2A−4+3C=−1
2(1+3C)−4+3C=−1
Simplify
More Steps
Evaluate
2(1+3C)−4+3C
Expand the expression
2+6C−4+3C
Subtract the numbers
−2+6C+3C
Add the terms
−2+9C
−2+9C=−1
Move the constant to the right-hand side and change its sign
9C=−1+2
Add the numbers
9C=1
Divide both sides
99C=91
Divide the numbers
C=91
Substitute the given value of C into the equation A=1+3C
A=1+3×91
Simplify the expression
A=1+3×9−1
Calculate
A=34
Substitute the given value of C into the equation B=2−C
B=2−91
Simplify the expression
B=2−9−1
Calculate
B=917
Calculate
⎩⎨⎧A=34B=917C=91
Solution
3(x−1)24+9(x−1)17+9(x+2)1
Show Solution
Find the roots
x1=−43+17,x2=4−3+17
Alternative Form
x1≈−1.780776,x2≈0.280776
Evaluate
((x−1)2(x+2))(2x2+3x−1)
To find the roots of the expression,set the expression equal to 0
((x−1)2(x+2))(2x2+3x−1)=0
Find the domain
More Steps
Evaluate
(x−1)2(x+2)=0
Apply the zero product property
{(x−1)2=0x+2=0
Solve the inequality
More Steps
Evaluate
(x−1)2=0
The only way a power can not be 0 is when the base not equals 0
x−1=0
Move the constant to the right side
x=0+1
Removing 0 doesn't change the value,so remove it from the expression
x=1
{x=1x+2=0
Solve the inequality
More Steps
Evaluate
x+2=0
Move the constant to the right side
x=0−2
Removing 0 doesn't change the value,so remove it from the expression
x=−2
{x=1x=−2
Find the intersection
x∈(−∞,−2)∪(−2,1)∪(1,+∞)
((x−1)2(x+2))(2x2+3x−1)=0,x∈(−∞,−2)∪(−2,1)∪(1,+∞)
Calculate
((x−1)2(x+2))(2x2+3x−1)=0
Remove the parentheses
((x−1)2(x+2))2x2+3x−1=0
Multiply the terms
(x−1)2(x+2)2x2+3x−1=0
Cross multiply
2x2+3x−1=(x−1)2(x+2)×0
Simplify the equation
2x2+3x−1=0
Substitute a=2,b=3 and c=−1 into the quadratic formula x=2a−b±b2−4ac
x=2×2−3±32−4×2(−1)
Simplify the expression
x=4−3±32−4×2(−1)
Simplify the expression
More Steps
Evaluate
32−4×2(−1)
Multiply
More Steps
Multiply the terms
4×2(−1)
Any expression multiplied by 1 remains the same
−4×2
Multiply the terms
−8
32−(−8)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
32+8
Evaluate the power
9+8
Add the numbers
17
x=4−3±17
Separate the equation into 2 possible cases
x=4−3+17x=4−3−17
Use b−a=−ba=−ba to rewrite the fraction
x=4−3+17x=−43+17
Check if the solution is in the defined range
x=4−3+17x=−43+17,x∈(−∞,−2)∪(−2,1)∪(1,+∞)
Find the intersection of the solution and the defined range
x=4−3+17x=−43+17
Solution
x1=−43+17,x2=4−3+17
Alternative Form
x1≈−1.780776,x2≈0.280776
Show Solution
