Help with some factoring

[Maitri22]

Hi newbie to forum,

Looking for help to understand the last step in factoring these 2 examples:
(A) 4x^2+18x-10

I understand how to factor this expression to get:
2(x(2x-1)+5(2x-1))
But I don’t understand the logic (how & why) you get to the answer:
=2(2x-1)(x+5)

Have the same problem with this one:

(B) 4x^2+12x+9... ....................fixed typo

I understand how to get to: (2x)^2 + 2(2x)(3) + 3^2
But don’t understand how it gets to:
=(2x+3)^2

Would appreciate if any could explain the how& why both these get to the final answer.
Thanks

 

[skeeter]

[MATH]2[x{\color{red}(2x-1)}+5{\color{red}(2x-1)}][/math]
common factor is [MATH]\color{red}(2x-1)[/MATH]
[MATH]2[{\color{red}(2x-1)}(x+5)][/MATH]

 

[Deleted member 4993]

Hi newbie to forum,

Looking for help to understand the last step in factoring these 2 examples:
(A) 4x^2+18x-10

I understand how to factor this expression to get:
2(x(2x-1)+5(2x-1))
But I don’t understand the logic (how & why) you get to the answer:
=2(2x-1)(x+5)

Have the same problem with this one:
(B) 4x^2+12+9
I understand how to get to: (2x)^+2(2x)(3)+3^2
But don’t understand how it gets to:
=(2x+3)^2

Would appreciate if any could explain the how& why both these get to the final answer.
Thanks

For (B) use:

(a + b)² = a² + 2*a*b + b²

 

[lookagain]

Have the same problem with this one:

Maitri22, your lines should look like this (as opposed to your typos):

(B) 4x^2 + 12x + 9
I understand how to get to: (2x)^2 + 2(2x)(3) + 3^2
But I don’t understand how it gets to:
= (2x + 3)^2

 

[Steven G]

If you had 3x+2x you would get (3+2)x = 5x.

If you had 3(2x-1) + 2(2x-1) = (3+2)(2x-1)

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

You have x (2x-1)'s and someone give you 5 more (2x-1)'s so now you have (x+5) (2x-1)'s which is denoted by (x+5)(2x-1).

 

[nasi112]

Maitri22, your lines should look like this (as opposed to your typos):

(B) 4x^2 + 12x + 9
I understand how to get to: (2x)^2 + 2(2x)(3) + 3^2
But I don’t understand how it gets to:
= (2x + 3)^2

it gets to [MATH](2x + 3)^2[/MATH] by completing the square

 

[HenryTorres]

As you saw above it (2x-1) is common in
2[x(2x−1)+5(2x−1)]
so
2[(2x-1)+(x+5)]

 

[lookagain]

As you saw above it (2x-1) is common in
2[x(2x−1)+5(2x−1)]
so
2[(2x-1)+(x+5)]

Henry Torres, that is wrong .

2[x(2x - 1) + 5(2x - 1)] =

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