Saturday

May 28, 2016
Posted by **Soly** on Monday, August 13, 2007 at 10:00pm.

Is the algebraic expression a polynomial? if it is write the polynomial in standard form,

1. 6x-9+8x^2 I got Yes; 8x^2+6x-9

Perform the indicated operations. Write the resulting polynomial in standard for.

3. (9x^5+20x^4+10) -(4x^5-10x^4-19) I got 5x^5+30x^4+29

Factor out the greatest common factor.

1. x^2(x-3)-(x-3) I got x^2(x-3)

Factor the trinomial, or state that the trinomial is prime.

3. x^2-4x-32 I got (x-4)(x-8)

4. 7x^2+39x+20 I got (7x+4)(x+5)

Factor the differnce of two squares.

7. 4x^2-49y^2 I got (2x+7y)(2x-7y)

Factor the perfect square trinomial.

10. x^2+10x+25 I got (x+5)^2

Factor using the formula for the sum or difference of two cubes.

13. x^3-8 This one I dont understand

14. 27x^3+64 This one I dont understand.

Factor completely, or state that the polynomial is prime.

21. 9x^4-9 This one I dont undertand

Factor and simplify the algebraci expression.

25. x^5/6 - x^1/6 This one I dont understand.

for

Factor out the greatest common factor.

1. x^2(x-3)-(x-3) I got x^2(x-3)

The greatest common factor would be (x-3)

The factored form would be:

(x-3)(x^2 - 1)

=(x-3)(x+1)(x-1)

"13. x^3-8 This one I dont understand

14. 27x^3+64 This one I dont understand."

For these two, there is an actual formula for the sum and difference of two cubes.

A^3 + b^3 = (A+B)(A^2 - AB + B^2) and

A^3 - b^3 = (A-B)(A^2 + AB + B^2)

so (x^3-8) = (x-2)(x^2+2x+4)

try the next one

"21. 9x^4-9 This one I dont undertand "

Factor out the 9 as a common factor, then you are left with a difference of squares

"25. x^5/6 - x^1/6 This one I dont understand. "

Tricky one.

How about taking out a common factor of x^(1/6)

x^5/6 - x^1/6

= x^(1/6)(x^(4/6) - 1)

= x^(1/6)(x^(2/3) - 1)

= x^(1/6)(x^(1/3) + 1)(x^(1/3) - 1)

(x^3-8) = (x-2)(x^2+2x+4) How did you come up with this? I get that you had to use the formula, but dont understand how you came up with the answer.

21. 9x^4-9= This is what I did for this one: 9(x^4-1)= (x^2+1)(x^2-1)= 9(x^2+1)(x^2-1).

25. x^5/6 - x^1/6 On this one, its a multiple choice question and x^1/6(x^1/3+1)(x-^1/3-1) is not a choice, but x^1/6(x^2/3-1 is a choice.

The choices are: a. x^5/6(1-x^2/3), b. x^1/6(x^5-1), c. x^1/6(x^2/3-1), d. x(x^2/3-1)

Sorry, my sister changed my screen name.

"(x^3-8) = (x-2)(x^2+2x+4) How did you come up with this? I get that you had to use the formula, but dont understand how you came up with the answer."

Look at the formula for

A^3 - B^3

Both of these must be perfect cubes.

The first factor is (A-B) in other words, the cube roots of those two terms

We had x^3 - 8

the cube root of x^3 is x of course, and the cube root of 8 is 2

For the second factor you just use A=x and B=2 to finish it.

for #21 you had

9(x^4-1)= (x^2+1)(x^2-1)= 9(x^2+1)(x^2-1).

the last part can be taken one more step, you have the difference of squares.

Final answer:9(x^4-1)= (x^2+1)(x^2-1)= 9(x^2+1)(x-1)(x+1).

for #25 I had x^1/6(x^2/3-1) as my second last line.

I simply took it one more step recognizing a difference of squares.

Since they allowed fractional exponents in their given answer, they should have allowed my final answer.

I need some help with some problems as well as need some checked. Could you please help me? Thanks!

1. How do you graph y=1/x

2. How do I the x and y intecepts using a graph?

Solve and check the linear equation.

9. (-4x-2)+7=-3(x+3) I got 14

10. -2[7x-7-6(x+1)]=2x+5 I got 21/4

Solve the equation.

13. (x+7)/4=2-(x-1)/6 I am not too sure how to do this one.

Find all the values of x satisfying the given conditions.

16. y1= (x+6)/3, y2=(x+8)/6, and y1=y2 This one I dont understand.

First write the value(s) that make the denominator(s) zero. Then solve the equation.

19. (x-8)/2x +5= (x+6)/x This one I also dont understand.

Determine whether the equation is an identity, a conditional equation, or inconsistent equation.

24. -2(x+7)+52=4x-6(x+3) I got Inconsistent

25. (3x+2)/4 +2= -7x/2 I got inconsistent