Question

Just a random math question.
Is expanding a trinominal a method of factoring?

Answer 1

yes, if it works

Answer 2

So in: 9x^2-24x+16, you could factor it by grouping and get (3x-4)(3x-4) or expand it and get (3x^2)-2(3*4)x+4^2?

Answer 3

@lizzylou169 its the same thing.... if given equation is a perfect square

Answer 4

Isn't 9x^2-24x+16 a perfect square?

Answer 5

yes

Answer 6

yes it is

Answer 7

But what if the function isn't a perfect square? What would I do then?

Answer 8

okay then depending upon the conditions expand it or factor it...

Answer 9

Okay then...I'll come up with a random function I suppose.

x^2+5x+7

Answer 10

this function does not have real values..

Answer 11

Math is frustrating. -__-

Answer 12

whatever ^-^ tell me you what do you wanna find??

Answer 13

I just want to see how to factor something by grouping and another method is the function is not a perfect square.

Answer 14

if*

Answer 15

let's take a function

x^2-5x+4=0 are you comfortable with this one?

Answer 16

Yep.

Answer 17

(x^2-5x)-(5x-4)
x(x-5)

There's no gcf in (5x-4)!

Answer 18

so now we first expand it
by means we modify our equation as

x^2-4x-x+4=0

why we did that??

because if you mulitply the last term and first term we get 4x^2 and we want our middle to split in the way so that product of the components of it would be 4x^2
here middle term is -5x (remember sign)
so we can write -5x=-4x-x and product of its components(that is of 4x and x ) is 4x^2
this is the method and only applicable to quadratic equation

so after expanding in this way we get

x^2-4x-x+4=0

now take x as a common then we get

x(x-4)-x-4=0 then take x-4 as common
we get

(x-1)(x-4)=0

so x-1=0
or x-4=0

so x=1 or x=4

Answer 19

@lizzylou169 hope it helps..

Answer 20

Thank you. :)

Answer 21

you are most welcome...!!