 Questions
 Algebra
I have a few more questions that I either need help with or just need checking.
Is the algebraic expression a polynomial? if it is write the polynomial in standard form,
1. 6x9+8x^2 I got Yes; 8x^2+6x9
Perform the indicated operations. Write the resulting polynomial in standard for.
3. (9x^5+20x^4+10) (4x^510x^419) I got 5x^5+30x^4+29
Factor out the greatest common factor.
1. x^2(x3)(x3) I got x^2(x3)
Factor the trinomial, or state that the trinomial is prime.
3. x^24x32 I got (x4)(x8)
4. 7x^2+39x+20 I got (7x+4)(x+5)
Factor the differnce of two squares.
7. 4x^249y^2 I got (2x+7y)(2x7y)
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^38 This one I don't understand
14. 27x^3+64 This one I don't understand.
Factor completely, or state that the polynomial is prime.
21. 9x^49 This one I don't undertand
Factor and simplify the algebraci expression.
25. x^5/6  x^1/6 This one I don't understand.
for
Factor out the greatest common factor.
1. x^2(x3)(x3) I got x^2(x3)
The greatest common factor would be (x3)
The factored form would be:
(x3)(x^2  1)
=(x3)(x+1)(x1)
"13. x^38 This one I don't understand
14. 27x^3+64 This one I don't 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 = (AB)(A^2 + AB + B^2)
so (x^38) = (x2)(x^2+2x+4)
try the next one
"21. 9x^49 This one I don't 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 don't 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^38) = (x2)(x^2+2x+4) How did you come up with this? I get that you had to use the formula, but don't understand how you came up with the answer.
21. 9x^49= This is what I did for this one: 9(x^41)= (x^2+1)(x^21)= 9(x^2+1)(x^21).
25. x^5/6  x^1/6 On this one, its a multiple choice question and x^1/6(x^1/3+1)(x^1/31) is not a choice, but x^1/6(x^2/31 is a choice.
The choices are: a. x^5/6(1x^2/3), b. x^1/6(x^51), c. x^1/6(x^2/31), d. x(x^2/31)
Sorry, my sister changed my screen name.
"(x^38) = (x2)(x^2+2x+4) How did you come up with this? I get that you had to use the formula, but don't 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 (AB) 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^41)= (x^2+1)(x^21)= 9(x^2+1)(x^21).
the last part can be taken one more step, you have the difference of squares.
Final answer:9(x^41)= (x^2+1)(x^21)= 9(x^2+1)(x1)(x+1).
for #25 I had x^1/6(x^2/31) 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. (4x2)+7=3(x+3) I got 14
10. 2[7x76(x+1)]=2x+5 I got 21/4
Solve the equation.
13. (x+7)/4=2(x1)/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 don't understand.
First write the value(s) that make the denominator(s) zero. Then solve the equation.
19. (x8)/2x +5= (x+6)/x This one I also don't understand.
Determine whether the equation is an identity, a conditional equation, or inconsistent equation.
24. 2(x+7)+52=4x6(x+3) I got Inconsistent
25. (3x+2)/4 +2= 7x/2 I got inconsistent
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