8/8/2019 A2T Ch9 Review

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Algebra II and TrigonometryChapter 9 Review Problems

1. Subtract: )2459()7325(2323 +!+!+!! x x x x x x

2. Multiply the following. Put your answers in standard form.

(a) )72)(65( +! mm (b) )245)(3( 2 +!! mmm

Review the following information about basic factoring.

The first step in any factoring problem is to factor out the GCF of the terms.

ex To factor 234 862 x x x !! , the first step is to factor out 22 x and write...

)43(2862 22234 !!"=!! x x x x x x

GCF "what's left"

The next step is to factor "what's left". For the final, the only types of factoringyou will need to do (after factoring out the GCF) are the following:

• Factorable Trinomials This is "normal factoring", like...

ex )1)(4(432 +!=!! x x x x

• Difference of Squares This always looks like ))((22 bababa +!=!

ex )32)(32(94 2 +!=! x x x

• Factoring by Grouping This is when you factor a cubic by pairing terms,factor out the GCF of each pair, and then use the distributive law.

ex To factor 1892 23 !!+ x x x :

)189()2(1892 2323 !!++=!!+ x x x x x x (Pair terms))2(9)2(2 +!++= x x x (Factor out GCF)

)2)(9( 2 +!= x x (Use the distributive law))2)(3)(3( ++!= x x x (Complete Factoring)

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3. Factor the following completely

(a) 234 12123 x x x +! (b) 542763 23 !!+ x x x (c) 5032 2 ! x

Review the following information about polynomial division.

We studing Long Division and Synthetic Division . Long Division is not on thefinal exam..Synthetic Division is. Synthetic Division only works if dividing apolynomial by a divisor that looks like k x ! or k x + .

ex. Divide 543 23 !+! x x x by 2! x

2 3 -4 1 -56 4 10

3 2 5 5

Put your coefficients on the top row. Besure to include 1's and 0's if necessary.First drop the 3 that is at the top left.Multiply by 2 to get 6. Then add down toget 2. Then Multiply by 2 to get 4, adddown to get 5, multiply by 2 to get 10 ...

The last number on the right is the remainder (5).The rest of the numbers on the bottom row are the coefficients of the quotient.So, for this division, we get...

R = 5 and Q = 523 2++ x x

4. Divide the following using synthetic division. Identify the quotient and remainder.

(a) 7557 23+!! x x x by 3+ x (b) 1123 3

+! x x by 2! x

Synthetic division is often the first step in a factoring problem.The typical chapter 9 question was :

ex Given that 3+ x is a factor of 7557 23+!! x x x (or given that –3 is a root)

factor 755723

+!! x x x completely.

Step1 Divide 7557 23+!! x x x by 3+ x .

You did this 4(a). You should have gotten: R=0 and Q = x 2 – 10x + 25.

So )2510)(3(7557 223 +!+=+!! x x x x x x

Step2 Factor the quadratic in the second parenthesis .

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)5)(5)(3(7557 23 !!+=+!! x x x x x x

5. Factor each of the following polynomials completely.

(a) 72185 23+!! x x x given that 3! x is one factor .

(b) 4524

+! x x given that 1 a root.

We also use synthetic division to find the zeros of a polynomial.The typical question is...

ex Given that 3+ x is a factor of 7557 23+!! x x x (or given that –3 is a root)

find all solutions to 07557 23=+!! x x x

Step1 This step is the same as above. Use synthetic division to write:

)2510)(3(7557 223 +!+=+!! x x x x x x

Step2 Either

• Factor the polynomial parenthesis to find its zeros.

)5)(5)(3(7557 23 !!+=+!! x x x x x x

0)5)(5)(3( =!!+ x x x gives 3!= x or 5

• Or…use the quadratic formula to find the zeros of the polynomial in parenthesis

6. Find all zeros (real and complex) of each polynomial below.

(a) 24269 23 !+! x x x given that 2 is a zero.

(b) 842 23 !+! x x x given that 2 is a zero

7. Sketch a graph of each equation below.

(a) 5)2( 4 +!= x y

(b) )6)(4)(2( !!!= x x x y

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(b) )6()4()2( 32 !!!= x x x y