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Basic Math Examples
y-1y2+4y+3+9y2+yy−1y2+4y+3+9y2+y
Step 1
Step 1.1
Factor y2+4y+3 using the AC method.
Step 1.1.1
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 3 and whose sum is 4.
1,3
Step 1.1.2
Write the factored form using these integers.
y-1(y+1)(y+3)+9y2+y
y-1(y+1)(y+3)+9y2+y
Step 1.2
Factor y out of y2+y.
Step 1.2.1
Factor y out of y2.
y-1(y+1)(y+3)+9y⋅y+y
Step 1.2.2
Raise y to the power of 1.
y-1(y+1)(y+3)+9y⋅y+y1
Step 1.2.3
Factor y out of y1.
y-1(y+1)(y+3)+9y⋅y+y⋅1
Step 1.2.4
Factor y out of y⋅y+y⋅1.
y-1(y+1)(y+3)+9y(y+1)
y-1(y+1)(y+3)+9y(y+1)
y-1(y+1)(y+3)+9y(y+1)
Step 2
To write y-1(y+1)(y+3) as a fraction with a common denominator, multiply by yy.
y-1(y+1)(y+3)⋅yy+9y(y+1)
Step 3
To write 9y(y+1) as a fraction with a common denominator, multiply by y+3y+3.
y-1(y+1)(y+3)⋅yy+9y(y+1)⋅y+3y+3
Step 4
Step 4.1
Multiply y-1(y+1)(y+3) by yy.
(y-1)y(y+1)(y+3)y+9y(y+1)⋅y+3y+3
Step 4.2
Multiply 9y(y+1) by y+3y+3.
(y-1)y(y+1)(y+3)y+9(y+3)y(y+1)(y+3)
Step 4.3
Reorder the factors of (y+1)(y+3)y.
(y-1)yy(y+1)(y+3)+9(y+3)y(y+1)(y+3)
(y-1)yy(y+1)(y+3)+9(y+3)y(y+1)(y+3)
Step 5
Combine the numerators over the common denominator.
(y-1)y+9(y+3)y(y+1)(y+3)
Step 6
Step 6.1
Apply the distributive property.
y⋅y-1y+9(y+3)y(y+1)(y+3)
Step 6.2
Multiply y by y.
y2-1y+9(y+3)y(y+1)(y+3)
Step 6.3
Rewrite -1y as -y.
y2-y+9(y+3)y(y+1)(y+3)
Step 6.4
Apply the distributive property.
y2-y+9y+9⋅3y(y+1)(y+3)
Step 6.5
Multiply 9 by 3.
y2-y+9y+27y(y+1)(y+3)
Step 6.6
Add -y and 9y.
y2+8y+27y(y+1)(y+3)
y2+8y+27y(y+1)(y+3)