Question

How do i factor 2x^2+11x+5

Answer 1

For simple integer factors as this one has, you want to find two values for the quadratic in the form ax^2+bx+c.   Let the two values be j and k.  These two values must satisfy two conditions.

jk=ac=10 and j+k=b=11, so j and k must be 1 and 10.

Now replace bx with jx and kx...

2x^2+x+10x+5  now factor 1st and 2nd pair of terms.

x(2x+1)+5(2x+1)

(x+5)(2x+1)

Answer 2

Factor by grouping.Tap for fewer steps...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⋅5=10a⋅c=2⋅5=10 and whose sum is b=−11b=-11.Tap for fewer steps...Factor −11-11 out of −11x-11x.2x2−11x+5=02x2-11x+5=0Rewrite −11-11 as −10-10 plus −1-12x2+(−10−1)x+5=02x2+-10-1x+5=0Apply the distributive property.2x2+(−10x−1x)+5=02x2+-10x-1x+5=0Remove parentheses.2x2−10x−1x+5=02x2-10x-1x+5=0Factor out the greatest common factor from each group.Tap for fewer steps...Group the first two terms and the last two terms.(2x2−10x)+(−1x+5)=02x2-10x+-1x+5=0Factor out the greatest common factor (GCF) from each group.2x(x−5)−1(x−5)=02xx-5-1x-5=0Factor the polynomial by factoring out the greatest common factor, x−5x-5.(x−5)(2x−1)=0x-52x-1=0Set x−5x-5 equal to 00 and solve for xx.Tap for fewer steps...Set the factor equal to 00.x−5=0x-5=0Since −5-5 does not contain the variable to solve for, move it to the right side of the equation by adding 55 to both sides.x=5x=5Set 2x−12x-1 equal to 00 and solve for xx.Tap for fewer steps...Set the factor equal to 00.2x−1=02x-1=0Since −1-1 does not contain the variable to solve for, move it to the right side of the equation by adding 11 to both sides.2x=12x=1Divide each term by 22 and simplify.Tap for more steps...x=12x=12The solution is the result of x−5=0x-5=0 and 2x−1=02x-1=0.x=5,12x=5,12