Any questions on the Section 5'5 homework

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Any questions on the Section 5.5 homework?
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Now please CLOSE YOUR LAPTOPS and turn off and
put away your cell phones.
Sample Problems Page Link (Dr. Bruce Johnston)
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Section 5.6

• Factoring Trinomials

NOTE Todays lecture again contains many
examples, so there probably wont be time for
homework in class after the lecture (and if we
dont get through all of the examples in class,
it will help you to look at the remaining
examples on the lecture slides before you start
the homework.)
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Section 5.6Factoring Trinomials

• Recall by using the FOIL method that
• (x 2)(x 4)
• x2 4x 2x 8
• x2 6x 8
• So to factor x2 6x 8 into (x __ ) (x __
  ), note that
• 6 is the sum of the two numbers 4 and 2 and 8 is
  the product of the two numbers.
• So well be looking for 2 numbers whose product
  is 8 and whose sum is 6.
• Note there are fewer choices for the product,
  so thats why we start there first.

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Factor the polynomial x2 13x 30.

• Since our two numbers must have a product of 30
  and a sum of 13, the two numbers must both be
  positive.
• Positive factors of 30 Sum of Factors
• 1, 30 31
• 2, 15 17

3, 10 13

• Note, there are other factors (like 65), but
  once we find a pair that works, we do not have to
  continue searching.
• So x2 13x 30 (x 3)(x 10).

Now check answer by multiplying the two factors
to see if you get back to the original trinomial.
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Factor the polynomial x2 11x 24.

• Since our two numbers must have a product of 24
  and a sum of -11, the two numbers must both be
  negative.
• Negative factors of 24 Sum of Factors
• -1, -24 -25
• -2, -12 -14

-3, -8 -11
So x2 11x 24 (x 3)(x 8). Now check
it!
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Factor the polynomial x2 2x 35.

• Since our two numbers must have a product of -35
  and a sum of -2, the two numbers will have to
  have different signs.
• Factors of -35 Sum of Factors
• -1, 35 34
• 1, -35 -34
• -5, 7 2

5, -7 -2
So x2 2x 35 (x 5)(x 7). Check it!
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Factor the polynomial x2 6x 10.

• Since our two numbers must have a product of 10
  and a sum of -6, the two numbers will have to
  both be negative.
• Negative factors of 10 Sum of Factors
• -1, -10 -11
• -2, -5 -7

Now we have a problem, because we have exhausted
all possible choices for the factors, but have
not found a pair whose sum is -6. So x2 6x 10
is not factorable and we call it a prime
polynomial.
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Factor the polynomial x2 11xy 30y2.

• We look for two terms whose product is 30x2y2 and
  whose sum is 11xy. The two terms will have to
  both be negative.
• Note each term will contain the variable y, for
  the sum to be 11y.
• Negative factors of 30x2y2 Sum of Factors
• -xy, -30xy -31xy
• -2xy, -15xy -17xy
• -3xy, -10xy -13xy

-5xy, -6xy -11xy
So x2 11xy 30y2 (x 5y)(x 6y).
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Factor the polynomial 3x6 30x5 72x4.

• First we factor out the GCF. (Always check for
  this first!)
• 3x6 30x5 72x4 3x4(x2 10x 24)
• Then we factor the trinomial.
• Positive factors of 24 Sum of Factors
• 1, 24 25
• 2, 12 14
• 3, 8 11

So 3x6 30x5 72x4 3x4(x2 10x 24)
3x4(x 4)(x 6).
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• You should always check your factoring results by
  multiplying the factored polynomial to verify
  that it is equal to the original polynomial.
• Many times you can detect computational errors or
  errors in the signs of your numbers (i.e those
  pesky dumb mistakes) by checking your results.

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What if the coefficient of the x2 term is not 1,
like it has been in all of the previous examples?
First, try to factor out a GCF. But what if
there isnt one, or if your leading coefficient
still ends up being larger than 1? Example
Factor 3x2 14x 8. There is no GCF that can
be factored out. However, this polynomial is NOT
prime. The factors of this trinomial are (3x
2)(x4). But how do we figure this out?
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Back to the problem Factor 3x2 14x 8. (Im
going to show you a method for these kinds of
problems that may be easier for you than the
guess and check method shown in the
textbook/online examples.) First multiply the
first and last coefficients together
38 24 Now look for two
factors of 24 that add up to 14 (the middle
number) 24 38, but 3 8 11 24
64, but 6 4 10 24 212, and 2 12
14, which is what were looking for. Now split
the middle term, 14x, into two pieces, 2x
12x This now gives 3x2 2x 12x 8 Now we
have a FOUR-TERM polynomial. What does this
suggest??
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You got it Factoring by grouping! 3x2 2x
12x 8 Factoring the first
pair gives 3x2 2x x(3x 2)
Factoring the 2nd pair gives 12x 8 4(3x
2) Putting it all back together now gives
x(3x2) 4(3x 2) (3x 2)(x4)
Done! Dont forget to always check your answer
by multiplying it back out. (You should get in
the habit of doing this on your homework so you
dont forget to do it on quizzes.)
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(When the coefficient of the x2 term is not 1)
Factor the polynomial 25x2 20x 4. First
multiply the first and last coefficients
together 254 100 Now find two factors of
100 that add up to 20 2 50? Nope (2 50
52) 520? Nope (5 20 25) 425?
Nope (4 2 5 29) 1010? Bingo! (10
10 20) Use this to split the middle term (20x)
into two pieces 25x2 10x 10x 4. Now
factor by grouping. Answer (5x 2)(5x 2)
(check it!) Whats a better way to write this?
(5x 2)2
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Factor the polynomial 21x2 41x 10. 1.
Multiply 2110210 2. Find 2 factors of 210 that
add up to -41 -10-21? -2-105? -5-42?
-15-14? -3-70? -6-35? Now factor by
grouping 21x2 - 6x 35x10 3x(7x 2) -5(7x
2) (7x 2)(3x 5) NOW CHECK IT!
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Factor the polynomial 3x2 20x 4. Step 1
3-4 -12 Step 2 possible factors of -12
3-4 sum -1 -34 sum 1 -112 sum
11 1-12 sum -11 -26 sum 4 2
-6 sum -4 There are no other possible ways to
factor -12, and none of the combinations add up
to 20, so this is a PRIME polynomial (it cant be
factored).
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Factor the polynomial 6x2y2 10xy2 4y2.
Remember to ALWAYS check for a GCF first. The GCF
of the three terms of this polynomial is 2y2, so
we factor that out first
6x2y2 10xy2 4y2 2y2(3x2 5x 2)
Now factor (3x2 -5x -2) by the factoring by
grouping method shown in the previous
examples. Final answer 2y2(3x 1)(x - 2)
(remember to check!)
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Reminder

• This homework assignment on Section 5.6 is due
• at the start of next class period.
• The tutor lab next door is open
• till 730 tonight.