Is there an easy way to factor when a≠1? Do Now: How can you factor when a=1? Can you still use the diamond problem when a≠1?

When can we use the regular diamond problem? When a=1 Ex: x2+x-20 =(x+5)(x-4) When we can factor out a constant so a=1 Ex: 4x2+20x+24 =4(x2+5x+6) =4(x+2)(x+3)

What can we do with quadratics when a≠1? When a≠1, there is more of a puzzle. We need to find the proper combination of numbers that we can add and multiply to form our quadratic. It is a big game!! To win this game we need to know the rules.

So what are the rules? First we need to make sure our equation is in standard form. Next we need to factor out any constants. If a=1, then we can use the regular diamond problem. If a≠1, then it starts to get interesting…

When a≠1… If a≠1, then we multiply the first and last coefficients to get our first clue. Ex. 5x2+11x+2=0 5*2=10 We take this clue and use it as the top number of a diamond problem with the normal bottom.

Now it gets interesting… We now have to factor our two numbers 10=10*1, 5*2 1=1*1 We are looking for a way to form the coefficients a and c using these factors. Note: We have to use both factors of a factor pair, we can’t mix and match. a=5=5*1 c=2=2*1

Last step Now that we have our factor pairs, we have to split them up. We take the ones we used to form a and put them with the x We take the ones we used to form c and put them by themselves. (5x+1)(x+2) Note: Make sure you break up the factors!

Example 6x2+11x+3

Try on your own 8x2+2x-15

One more 9x2-18x+8

Review Change the diamond problem Multiply first and last terms Factor, look for factor pair that will sum to the middle term Break down those factors, look for two pairs that we can use to form the first and last terms. Write as two binomials, make sure you break up the factor pairs.

Summary/HW Why do we need to change the diamond problem when a≠1? HW: pg 76, 1-6