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Algebra Examples
2x2+x-15=02x2+x−15=0
Step 1
Step 1.1
For a polynomial of the form ax2+bx+cax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=2⋅-15=-30a⋅c=2⋅−15=−30 and whose sum is b=1b=1.
Step 1.1.1
Multiply by 11.
2x2+1x-15=02x2+1x−15=0
Step 1.1.2
Rewrite 11 as -5−5 plus 66
2x2+(-5+6)x-15=02x2+(−5+6)x−15=0
Step 1.1.3
Apply the distributive property.
2x2-5x+6x-15=02x2−5x+6x−15=0
2x2-5x+6x-15=02x2−5x+6x−15=0
Step 1.2
Factor out the greatest common factor from each group.
Step 1.2.1
Group the first two terms and the last two terms.
(2x2-5x)+6x-15=0(2x2−5x)+6x−15=0
Step 1.2.2
Factor out the greatest common factor (GCF) from each group.
x(2x-5)+3(2x-5)=0
x(2x-5)+3(2x-5)=0
Step 1.3
Factor the polynomial by factoring out the greatest common factor, 2x-5.
(2x-5)(x+3)=0
(2x-5)(x+3)=0
Step 2
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
2x-5=0
x+3=0
Step 3
Step 3.1
Set 2x-5 equal to 0.
2x-5=0
Step 3.2
Solve 2x-5=0 for x.
Step 3.2.1
Add 5 to both sides of the equation.
2x=5
Step 3.2.2
Divide each term in 2x=5 by 2 and simplify.
Step 3.2.2.1
Divide each term in 2x=5 by 2.
2x2=52
Step 3.2.2.2
Simplify the left side.
Step 3.2.2.2.1
Cancel the common factor of 2.
Step 3.2.2.2.1.1
Cancel the common factor.
2x2=52
Step 3.2.2.2.1.2
Divide x by 1.
x=52
x=52
x=52
x=52
x=52
x=52
Step 4
Step 4.1
Set x+3 equal to 0.
x+3=0
Step 4.2
Subtract 3 from both sides of the equation.
x=-3
x=-3
Step 5
The final solution is all the values that make (2x-5)(x+3)=0 true.
x=52,-3