how do you factor trinomials!!?

Okay, start off by looking for common factors

4k^2 + 48k + 108

= 4(k^2 +12k + 27)

Now we need to check out the inside the brackets...

We know that you can get 27 only by multiplying 3 and 9

3 + 9 = 12

so we can go to

= 4(k+3)(k+9)

and that's it finished

why did we look at "3 + 9"

Well, think of what happens when you multiply out

(k + a)(k + b) where a and b are two numbers that are constant

= k^2 +ak + bk + ab

= k^2 + (a + b)k + ab

Note that 3 and 9 satisfy a+ b = 12 and ab = 27

k, by the way, is called the variable...

Now, let's look at the simplest one (the last one)

9 is a factor of 54

x is a factor of x^2 and x

so

9x^2 + 54x

= 9x(x + 6)

Now, back to the second one:

12r^2 - 74r + 112

.... 2 is a factor of each term

2(6r^2 - 37r + 56)

What we need to look for is multiples of 6 and 56 which add up to 37

2 and 198 --> 200

3 and 112 --> 115

4 and 84 --> 88

6 and 56 --> 62

7 and 48 --> 55

8 and 42 --> 50

14 and 24 --> 38

21 and 16 --> 37

SO we can break the expression into

2(6r^2 - 16r - 21r + 56)

16 = 2*8 (the 2 goes in one factor(in front of an r) and the eight into the other)

21 = 3*7 (the 3 goes in one factor (in front of an r) and the 7 goes into the other)

==>2(2r - 7)(3r - 8)

and we are done (after a while, you start to see what will work and what doesn't, but that takes practice)

and the remaining one:

You could do it the same way as the last one, except that you'd look at the DIFFERENCE between the factors (look at the sign in front of the constant term -- it's negative, so look for the difference; if positive, look for the sum)

Or...

9x^2 -18x - 40

= 9(x^2 -2x) - 40 ... complete the square

= 9(x^2 - 2x + 1) - 9 - 40

= 9(x-1)^2 - 49 ... difference of squares

= (3(x-1) +7)(3(x-1)-7)

= (3x -3 + 7)(3x -3 - 7)

= (3x + 4)(3x -10)

6 and 56 --> 62, so this does not work

12 and 28 --> 40, so this does not work

Hi, I think she is right. The best is to probably use a tutor, but I have a cool tip i think you might like.

You can actually get tuition help for free, lol. There is this site i found called http://bloomstutors.com.au/ you can post questions like on here but instead, tutors answer them. So post it under maths and you can have them do the homework for you. :)

4 ( k^2 + 12k + 27)

4 ( k + 3 ) ( k + 9 )