Transcript Is there an easy way to factor when a≠1?

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