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. 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 don't understand
14. 27x^3+64 This one I don't understand.
Factor completely, or state that the polynomial is prime.
21. 9x^4-9 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.
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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 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 = (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 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^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 don't 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 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 (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 don't 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 don't 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
4 answers