OpenStudy (anonymous):

*~*~*~*HELP PLEASE*~*~*~* Factor 400x^2 + 160x + 16 completely.

5 years ago
OpenStudy (anonymous):

Oh.. and I have to factor it using a special product or something?

5 years ago
OpenStudy (anonymous):

a=400, b=160, c=16 ac=6400 We need to find factors of 6400 that add up to 160 do start the problem.

5 years ago
OpenStudy (anonymous):

80 and 80

5 years ago
OpenStudy (anonymous):

Ok. Now we will form this square:\[\left[\begin{matrix}400x^{2} & 80x \\ 80x & 16\end{matrix}\right]\]Notice that ax^2 is the first element of the first row, c is the second element of the second row, and the two factors times x are the other two entries (it doesn't matter which spot you put the factors in). Now, the next step is to find the greatest common factors of each row and each column. For example, the row with 80x and 16 has 16 as its greatest common factor. (In your work write this in front of the that row. I would show you exactly what I mean, but I don't think the formatting here will let me do it.) Then do that with the rest of the rows and columns. Finally, take those factors that you found and they will be the factored solution. The factors for the columns will be one set of parenthesis, and the factors for the rows will be the other set.

5 years ago
OpenStudy (anonymous):

400x^2 + 160x + 16 = (20x)^2 +160x+4^2 = (20x)^2 +2(20x)(4)+4^2 can you do it now??

5 years ago
OpenStudy (anonymous):

hint: a^2+2ab+b^2 = (a+b)^2 = (a+b)(a+b)

5 years ago
OpenStudy (anonymous):

The only other thing to expand it to a more general case (where you would say have negative factors or a is negative) is that the factor you write in front of the row or above the column has the same sign as the closest element to it. So for example if a was negative, the factor for that row would be negative, regardless of whether or not the other entry in that row was negative or positive. The same holds true for that column.

5 years ago
OpenStudy (anonymous):

@Aylin I don't understand the square thing with the ax^2 and c? and @rizwan_uet where would you put the factors that add up to 160?

5 years ago
OpenStudy (anonymous):

That's just a box you put the entries into to help with factoring.

5 years ago
OpenStudy (anonymous):

oh ohk thanks

5 years ago
OpenStudy (anonymous):

When you have your final answer you can post it here and either I or rizwan should be able to tell you if it is correct or give some more help if needed.

5 years ago
OpenStudy (anonymous):

ohk i will(: Thanks guys I wish I could pick 2 best answers

5 years ago
OpenStudy (anonymous):

I got (20x + 4)^2

5 years ago
OpenStudy (anonymous):

That does work, but it is not the simplest form. 20x and 4 both share a factor of 4. Do you know how to make it simpler?

5 years ago
OpenStudy (anonymous):

so it would be 5x + 1?

5 years ago
OpenStudy (anonymous):

Sort of. \[(20x+4)^{2}=(4(5x+1))^{2}=16(5x+1)^{2}\]

5 years ago
OpenStudy (anonymous):

oh ohk thank you!!

5 years ago
OpenStudy (anonymous):

you have this ok (20x)^2 +160x+4^2 now from the factors (20x)^2 and 4^2 i can write (20x)^2 +2(20x)(4)+4^2 because i know the formula a^2+2ab+b^2 = (a+b)^2 thus (20x)^2 +2(20x)(4)+4^2 = (20x+4)^2

5 years ago
OpenStudy (anonymous):

now (20x+4)^2 = (20x+4)(20x+4) thus the required factors are (20x+4)(20x+4)

5 years ago
OpenStudy (anonymous):

oh I understand now thank you so much!

5 years ago
OpenStudy (anonymous):

welcome

5 years ago