Factoring Polynomials Grouping, Trinomials, Binomials, GCF,Quadratic form & Solving Equations.

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Factoring Polynomials Grouping, Trinomials, Binomials, GCF ,Quadratic form & Solving Equations

Transcript of Factoring Polynomials Grouping, Trinomials, Binomials, GCF,Quadratic form & Solving Equations.

Factoring Polynomials

Grouping, Trinomials, Binomials, GCF ,Quadratic form & Solving Equations

Student will be able to Factor by Grouping terms

When polynomials contain four terms, it is sometimes easier to group like terms in order to factor.

Your goal is to create a common factor. You can also move terms around in the

polynomial to create a common factor. Practice makes you better in recognizing

common factors.

Factoring Four Term Polynomials

Do now: find the GCf of the first two terms and the last two terms:

3x3 −12x2 −6x+24

3x2 and 6

Group together and Factor each one separately:

They share a common factor of (x-4)

Write 2 factors:

Write the common factor once and put the outside terms together:

(3x3 −12x2 )−(6x+24)

3x2 (x−4)−6(x−4)

(3x2 −6)(x−4)

Factor by GroupingExample 1:

FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy – 21y Factor the last two terms: + 5x - 35 =

Factor by GroupingExample 1:

FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy - 21y = 3y (x – 7) Factor the last two terms: + 5x - 35 = 5 (x – 7) The green parentheses are the same so

it’s the common factor

Factor by GroupingExample 1:

FACTOR: 3xy - 21y + 5x – 35 Factor the first two terms: 3xy - 21y = 3y (x – 7) Factor the last two terms: + 5x - 35 = 5 (x – 7) The green parentheses are the same so it’s the

common factor Now you have a common factor

(x - 7) (3y + 5)

Factor by Grouping Example 2:

Factor by Grouping Example 2: FACTOR: 6mx – 4m + 3rx – 2r Factor the first two terms: 6mx – 4m = 2m (3x - 2) Factor the last two terms: + 3rx – 2r = r (3x - 2) The green parentheses are the same so it’s the

common factor Now you have a common factor

(3x - 2) (2m + r)

Factor by Grouping Example 3:

FACTOR: y3– 5y2 -4y +20

Factor by Grouping Example 3:

FACTOR: y3– 5y2 - 4y +20 Factor the first two terms: y3– 5y2 = y2 (y - 5) Factor the last two terms: - 4y +20 = -4 (y – 5) The green parentheses are the same! y2 (y - 5) and -4 (y - 5) Now you have the difference of two squares! look at red ( ): (y - 5) (y2 - 4) : answer: (y - 5) (y - 2) (y + 2)

See worksheet “Factor by grouping” Try first 4 problems.

Using Factor by Grouping to solve a polynomial function:From the last example, suppose it was an equation…..

y3– 5y2 - 4y +20 = 0 (y - 5) (y - 2) (y + 2) = 0 y=5 y = 2 y=-2So the solution set is { 5,2,-2}

Factor first, then set factors = 0

3x3 −12x2 −6x+24 =0

(3x2 −6)(x−4)=0

3x2 −6 =0 X-4=0

solve

3x3 −12x2 −6x+24 =0

(3x2 −6)(x−4)=0

3x2 −6 =0

3x2 =6

x2 =2

x=± 2

X=4

X-4=0

{4, 2,− 2}

Hand this one in:

Solve for all roots:

3x3 - 4x2 -27x +36 = 0

Factoring Trinomials

Factoring Trinominals

1. When trinomials have a degree of “2”, they are known as quadratics.

2. We learned earlier to use the last term’s factors to factor trinomials that had a “1” in front of the squared term.

x2 + 12x + 35

So… 7 and 5 or 35 and 1

Factoring Trinominals

1. When trinomials have a degree of “2”, they are known as quadratics.

2. We learned earlier to use the last term’s factors to factor trinomials that had a “1” in front of the squared term.

x2 + 12x + 35

(x + 7)(x + 5)

Because 7 + 5 = 12!

More Factoring Trinomials

3. When there is a coefficient larger than “1” in front of the squared term, we can use a method we will call, the “am” add, multiply method to find the factors.

3. Always remember to look for a GCF before you do ANY other factoring.

More Factoring Trinomials

5. Let’s try this example3x2 + 13x + 4

(3x )(x )

Write the factors of the last term…1,4 2,2Multiply using foil until you get the middle

term of the trinomial. If so, you’re done!

More Factoring Trinomials

3x2 + 13x + 4 (3x + 1 )(x + 4 )3x2 + 12x + 1x + 4= 3x2 + 13x + 4 ✓

Difference of Squares

Difference of Squares When factoring using a difference of squares,

look for the following three things:– only 2 terms– minus sign between them– both terms must be perfect squares – No common factors

If all of the above are true, write two ( ), one with a + sign and one with a – sign :

( + ) ( - ).

Try These, (if possible)

1. a2 – 16 2. x2 – 25 3. 4y2 – 16 4. 9y2 – 25 5. 3r2 – 81 6. 2a2 + 16

answers:

1. a2 – 16 (a + 4) (a – 4) 2. x2 – 25 (x + 5) (x – 5) 3. 4y2 – 9 (2y + 3) (2y – 3) 4. 9y2 – 25 (3y + 5) (3y – 5) 5. 3r2 – 81 *3 is not a square! 6. a2 + 16 Not a difference!

Perfect Square Trinomials

Perfect Square Trinomials When factoring using perfect square

trinomials, look for the following three things:– 3 terms– last term must be positive– first and last terms must be perfect

squares If all three of the above are true, write one (

)2 using the sign of the middle term.

Try These

1. a2 – 8a + 16 2. x2 + 10x + 25 3. 4y2 + 16y + 16 4. 9y2 + 30y + 25

Factoring Completely

Factoring Completely Now that we’ve learned all the types of

factoring, we need to remember to use them all.

Whenever it says to factor, you must break down the expression into the smallest possible

factors.

Let’s review all the ways to factor.

Types of Factoring1. Look for GCF first.2. Count the number of terms:

a) 4 terms – factor by groupingb) 3 terms -

1. look for perfect square trinomial2. if not, try “am” method

c) 2 terms - look for difference of squares

If any ( ) still has an exponent of 2 or more, see if you can factor again.

These may take 2 steps!

1. 3r2 – 18r + 27 2. 2a2 + 8a - 8

Answers:

1. 3r2 – 30r + 27 3(r2 - 10r + 9) 3(r – 9) (r – 1)

2. 2a2 + 8a – 8 2(a2 + 4a – 4)

Solving Equations by FactoringCompletely

1) x5 −4x2

2) 3x2 −18x+27 =0

Do Now:1)Factor completely2)Solve for x

Steps to Solve Equations by FactoringCompletely

set each factor = 0 and solve for the unknown.

x3 + 12x2 = 0 1. Factor GCF

x2 (x + 12) = 0

Steps to Solve Equations by FactoringCompletely

set each factor = 0 and solve for the unknown.

x3 + 12x2 = 0 1. Factor GCF

x2 (x + 12) = 0 2. (set each factor = 0, & solve)

x2 = 0 x + 12 = 0

x=0 x = -12

You now have 2 answers, x = 0 and x = -12

Factor completely:

x3 + x2 −12x=0

Factor completely:

x3 + x2 −12x=0

x(x2 + x−12)=0x(x+ 4)(x−3)=0

0 , −4 , 3X =

Solving higher degree functions

Quadratic form: ax4 + bx2 + c = 0 Example: x4 +2x2 -24 = 0 Factor: (x2 )(x2 )=0

Solving higher degree functions

Quadratic form: ax4 + bx2 + c = 0 Example: x4 +2x2 -24 = 0 Factor: (x2 +6 )(x2 – 4 ) = 0 x2 +6=0 x2 – 4 =0 x2 =-6 x2 = 4 x = 2, -2 x =±i 6

Try this one:

X4 – 13x2 +36 = 0

Factor first:

X4 – 13x2 +36 = 0

(x2 – 9)(x2 – 4)=0

Solutions:

X4 – 13x2 +36 = 0

This one can be verified on the calculator.

X = 2,-2,3,-3

(x2 – 9)(x2 – 4)X2-9=0 x2-4=0X=3,-3 x= 2,-2

Hand this one in

Quadratic form:

x4 +6x2 −27 =0

Ans:

X=

x4 +6x2 −27 =0

(x2 +9)(x2 −3)=0

x2 +9 =0 x2 −3=0

x2 =−9 , x2 =3

x=3i,−3i , x= 3,− 3