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23. Solve the equation. Check for extraneous solutions. 9|9 – 8x| = 2x + 3 9l9-8xl=2x+3 subract 8 9l9l=-6x+3 subratc 3 9l6l=-6x 54=-6x -9.33333333
13. What are the real or imaginary solutions of the polynomial equation? 27x^3 + 125 = 0 27x^3+125=0 27x^3=–125x^3=–125/27x^3=(–5/3)^3x=–5/3
|dw:1357254686057:dw|

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18. Use Pascal’s Triangle to expand the binomial. ( d – 5)^6 =(d^4-10d+25d-10d+100d-250d+25d-250d+625)(d-10d+25)=(d^4-20d+150d-500d+625)(d-10d+25)=d^6-10d^5+25d^4-20d^5+200d^4-500d+150d^4-1500d+3750d-500d+5000d-12500d+625d-6250d+15625=d^6-30d^5+375d^4-2500d+9375d-18750d+15625 21. Explain how to find the degree of a polynomial. To find the degree of a polynomial its the same number term of highest degree. for an example, if u have terms 5 10 15 20, 20 would be the degree.
your process is incorrect for the second one, you will need to take the cube root of the fraction
these are questions i missed on previous tests & id like to correct them
|dw:1357254904914:dw|ok still working on the same problem
|dw:1357254993499:dw|
\[27x^3 + 125 = 0\] factor the sum of two cubes as \[(3x+5)(3x^2-15x+25)\] first one gives you your answer, quadratic formula gives the answer for the two complex roots of \(3x^2-15x+25\)
|dw:1357255010162:dw|
For your first problem (# 23), you will have 2 solutions because of the absolute value expression. Case #1: For x > 9/8 9(8x - 9) = 2x + 3 -> x = 1.2 Case #2: For x <= 9/8 9(9 - 8x) = 2x + 3 -> x = 39/37
For #13, did you use the \[(a +b) ( a^{2} + ab + b ^{2})\] method?
*correction (a+b) (a^2 - ab + b^2)
@satellite73 please check your factoring
oh yeah, it is wrong isn't it?
\[27x^3+125=(3x+5)(9x^2-15x+25)\] is more like it
thx, @satellite73 you saved me a bunch of typing. Great approach, explanation, and theory. Just a typo, as we all do that from time-to-time.

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