How to factor the quadratic function from the general form!

[honest_denverco09]

I have a question about how can i use the "rectangular diagram" to convert the general form of quadratic function to the square (Factor the quadratic function) :
Actually, i don't know how to find the middle term in the trinomial. For instant :

x^2 - 22x + 121
so we have

But i don't know how can i find the "a" and "b" . And i know that a + b = -22 and ab = 121 .

[Martingale]

I have a question about how can i use the "rectangular diagram" to convert the general form of quadratic function to the square (Factor the quadratic function) :
Actually, i don't know how to find the middle term in the trinomial. For instant :

x^2 - 22x + 121
so we have

But i don't know how can i find the "a" and "b" . And i know that a + b = -22 and ab = 121 .

What are the factors of 121?

[stapel_eliz]

Actually, i don't know how to find the middle term in the trinomial. For instant :

x^2 - 22x + 121

Following from the process demonstrated here, look at the constant term: +121. Since the constant is postive, and since you're multiplying to get this value, then you multiplied either two positive or two negative numbers.

Now look at the coefficient of the middle term: -22. Since this value is negative, and since you're adding to get this value, then you must have multiplied two negative numbers.

So you're looking for a pair of factors of 121 which, when you put "minus" signs on them, sum to -22. The factor pairs are:

. . . . .1 and 121
. . . . .11 and 11

Which pair works?

[power_up_plus]

Actually, i don't know how to find the middle term in the trinomial. For instant : x^2 - 22x + 121

1st:
check to see if the squared term has 1 as the coefficent

2nd:
set up the prentheis (x + or - __)(x + or - __)

3rd:
find two factors of 121 that when added together equal -22 (in this case are -11 and -11)

4th:
place the two numbers into the blank spots above

Answer:
x^2 - 22x + 121 = (x - 11)(x - 11)

Good Luck

[honest_denverco09]

Thank you for your answering, but it is still difficult to know what are the factor of some vast number. Here is the other way to find the middle term of trinomial function (Is my problem is to find the letter "a" and "b").

We have : x^2 - 22x + 121 = y

First Step : Identify the c in form Ax^2 + Bx + C = y . So C = 121
Second Step : Square root of C = the "a" = the "b" . Because Square root of 121 is 11, so a = b = 11
Third Step : Refer the letter b to form the sign of a and b to Make sure that a + b = B
. Because B = -22 (negative), so a = b = -11
Last Step :Put a and b into factor form (x - a)(x - b) = (x - a)^2 . We have y = (x + 11)(x - 11) = (x - 11)^2

Remember that the letter "a" and "b" are equal.

[stapel_eliz]

Thank you for your answering, but it is still difficult to know what are the factor of some vast number.

Which is why the big numbers are almost always easily factorable, such as 11² = 121!

[dylanr]

well you have the equation x^2-22x+121=y

First: You find factors of 121 and that add up to -22.
11 and 11 are one, and 121 and 1 are anoher one. 11 and 11 are the only ones that are factors of 121 and add up to 22.

Second: Set up parenthisis with X in both- (x+/-_) (x+/-_)

Third: Insert 11 and 11 into the equation so they add up to -22.

Answer: (x-11) (x-11)

[allie]

Ok, to factor x^2-22x+121=y .
First, set up two separate parenthesis (x-_)(x-_)
* both of the signs would be a subtraction sign
Next, find two factors of 121 that you can subtract together to equal a -22
* a negative 11 multiplied by another negative 11 would equal a positive 121, subtract the two factors (-11 & -11) and it equals a -22 for your middle term
So, your final factored equation would be, (x-11)(x-11)