Verified answer

It's (2x+1)(x+5)

because if you multiply it together you get:

First 2x * x = 2x²

Outside 2x*5 = 10x

Inside 1*x=x

Last 1*5=5

= 2x² + 10x + x + 5

= 2x² + 11x + 5

So you know the values before x in the bracket need to multiply to make the value in front of x² in your original equation. The only multiples of 2 are 1 and 2 so they go in the brackets.

(2x )(x )

You know the values on the end must multiply to equal 5, so you need 1 and 5 on the end.

But if you put it in like this:

(2x + 5)(x + 1)

then 5*x= 5x and 2x*1 = 2x and 5x+2x = 7x which isn't the value in the original equation.

So if you swap them around:

(2x + 1)(x + 5) it's 1*x = x and 2x*5=10x and x+10x = 11x which is in the equation so it's correct.

2 x*x + 11x + 5

= 2 x*x + 10x + x + 5

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

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

Hope this helps you to do your Hw.

This is known as factorising by middle-term factorisation.

i actually just learned this last week.

since you cant factor out 2x you have to include it in your bracketts like this:

(2x_)(x_) and the blanks would be where the numbers go. and since you have to include the 2 inside the brackets, the rule that the multiple of 5 will add up to 11 will not work. you yust have to try different combos. this is easy because there is only one multiple of 5, 5 and 1. you have to put them in.

once yo put them in you multipy the inner and outers and put them together and it should equal 11x

it will be (2x+)(x+5) because 2x multiplied by 5 (the outers) is 10x and x times 1(the inners)bis 1x and 10x +1x is 11x.

sorry its a little confusing but good luck

you could also use abc method which is easier. You mutliply the first number (2) to the last (5) then you get ten. So now you want to numbers that mutliply to ten but add up to eleven...so here it goes

10x1=10

10+1=11

so you plug in 10 and 11

2x² + 10x + 1x +5

then you factor by groups

(2x² +10x) ( 1x+5)

so now you find GCF

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

so its not now

(2x+1) (x + 5)

sorry if i confused its just easier to me this way but longer.

(2x + 1) (x + 5 )

Sometimes you will have a coefficient in front of a variable raised to some power. Just the nature of the beast. Easily tamed with a couple o' whacks of the math stick.

(2x+1)(x+5)

wolframalpha.com will do the work for you.

http://www.wolframalpha.com/input/?i=2x%C2%B2+%2B+...