# I Tried So HArd But I Needed Help!?

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03.06 Factoring by Grouping Algebra 2

Question 1 (Multiple Choice Worth 1 points)

Select one of the factors of 4x^2 + 5x – 6

a (x – 3)

b (4x – 3)

c (4x + 2)

d (x + 6)

Question 2 (Multiple Choice Worth 1 points)

Select one of the factors of x^3y^2 + 8xy^2 – 5x^2 – 40

a (xy^2 + 5)

b (x^2 + 4)

c (xy^2 – 5)

d (x^2 – 8)

Question 3 (Multiple Choice Worth 1 points)

Select one of the factors of 3x^2 + 10x + 3

a (3x + 1)

b (3x + 3)

c (3x – 1)

d None of the above

Question 4 (Multiple Choice Worth 1 points)

Select one of the factors of x^3 + 4x^2 + 9x + 36.

a (x^2 + 4)

b (x^2 + 9)

c (x + 9)

d None of the above

Question 5 (Essay Worth 3 points)

Explain, in complete sentences, how you would completely factor 3x^3 + 27x^2y – 24xy – 216y^2 and check your factors for accuracy.

Question 6 (Essay Worth 3 points)

Explain, in complete sentences, how you would completely factor 20x^2 – 28x – 48 and check your factors for accuracy.

• 1) Remember that (ax+b)(cx+d) = ac + (ad + bc)x + bd. So if (x-3) was a factor of 4x^2 + 5x – 6, then the other factor would have to be (4x+2) in order to get “4x^2” for the first term and “-6” for the last term. However, (x-3)(4x+2) = 4x^2 – 12x + 2x – 6 = 4x^2 – 10x – 6. This gives us -10x in the middle, not 5x. So the answer can’t be x-3. Try the same strategy with the other choices.

• 4x^2 + 5x – 6

by grouping

I would say screw grouping

but that might offend people

so I won’t

use the quadratic formula to find roots

give the answer as grouping

b^2-4ac

25+96

121

+/-11

rest of q.f.

(-b+/-11)/2a

(-5+/-11)/8

6/8, -16/8

3/4, -2

x=3/4

4x=3

4x-3=0

(4x-3)

x= -2

x+2=0

(x+2)

the answer in ‘grouping’ would be

4x(x+2)-3(x+2)

or

x(4x-3)+2(4x-3)

one more

Select one of the factors of x^3y^2 + 8xy^2 – 5x^2 – 40

think what the other factor might need to be to satisfy the first and last term x^3y^2 and -40

a (xy^2 + 5)

needs to be (x-8)

b (x^2 + 4)

needs to be (xy^2-10)

c (xy^2 – 5)

needs to be (x^2+8)

d (x^2 – 8)

needs to be (xy^2+5)

if you multiply them out

c is the one

c.

[re-post the other questions 2 at a time,k?]

;~)

• Q1

(4x – 3)(x + 2) = 4x^2 + 8x – 3x – 6 = 4x^2 + 5x – 6

(x+6)(4x – 1) = 4x – x + 6x – 6 = 4x + 5x – 6

• 1.)

4x^(2)+5x-6

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-24) that add up to b (5).

a=4, b=5, c=-6

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-24) that add up to b (5).In this problem 2*-(3)/(4)=-(6)/(4) (which is (c)/(a)) and 2-(3)/(4)=(5)/(4) (which is ((b)/(a)) , so insert 2 as the right hand term of one factor and -(3)/(4) as the right-hand term of the other factor.

(x+2)(x-(3)/(4))

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.

(x+2)(4x-3)

2.)

x^(3)y^(2)+8xy^(2)-5x^(2)-40

Factor the greatest common factor (GCF) from each group.

(xy^(2)(x^(2)+8)-5(x^(2)+8))

Factor the polynomial by grouping the first two terms together and finding the greatest common factor (GCF). Next, group the second two terms together and find the GCF. Since both groups contain the factor (x^(2)+8), they can be combined.

(xy^(2)-5)(x^(2)+8)

3.)

3x^(2)+10x+3

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (9) that add up to b (10).

a=3, b=10, c=3

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (9) that add up to b (10).In this problem 3*(1)/(3)=(3)/(3) (which is (c)/(a)) and 3+(1)/(3)=(10)/(3) (which is ((b)/(a)) , so insert 3 as the right hand term of one factor and (1)/(3) as the right-hand term of the other factor.

(x+3)(x+(1)/(3))

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.

(x+3)(3x+1)

4.)

x^(3)+4x^(2)+9x+36

Factor the greatest common factor (GCF) from each group.

(x^(2)(x+4)+9(x+4))

Factor the polynomial by grouping the first two terms together and finding the greatest common factor (GCF). Next, group the second two terms together and find the GCF. Since both groups contain the factor (x+4), they can be combined.

(x^(2)+9)(x+4)

5.)

3x^(3)+27x^(2)y-24xy-216y^(2)

Factor out the GCF of 3 from each term in the polynomial.

3(x^(3))+3(9x^(2)y)+3(-8xy)+3(-72y^(2))

Factor out the GCF of 3 from 3x^(3)+27x^(2)y-24xy-216y^(2).

3(x^(3)+9x^(2)y-8xy-72y^(2))

Factor the greatest common factor (GCF) from each group.

3(x^(2)(x+9y)-8y(x+9y))

Factor the polynomial by grouping the first two terms together and finding the greatest common factor (GCF). Next, group the second two terms together and find the GCF. Since both groups contain the factor (x+9y), they can be combined.

3(x^(2)-8y)(x+9y)

6.)

20x^(2)-28x-48

Factor out the GCF of 4 from each term in the polynomial.

4(5x^(2))+4(-7x)+4(-12)

Factor out the GCF of 4 from 20x^(2)-28x-48.

4(5x^(2)-7x-12)

For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (-60) that add up to b (-7).In this problem 1*-(12)/(5)=-(12)/(5) (which is (c)/(a)) and 1-(12)/(5)=-(7)/(5) (which is ((b)/(a)) , so insert 1 as the right hand term of one factor and -(12)/(5) as the right-hand term of the other factor.

4(x+1)(x-(12)/(5))

Remove the fraction by multiplying the first term of the factor by the denominator of the second term.

4(x+1)(5x-12)

• I think you should add six more sub-questions and three more essays to your original question and you’ll get even more answers.

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