Factorization

Question

Factor the expression.
50k cubed-40k squared+75k-60
Anyone know the answer?

 

Solution

50k^3 -40k^2 +75k -60
=5(10k^3 -8k^2 +15k -12)
=5k(2k(5k-4) +3(5k-4))
=5k(2k + 3) (5k -4)

Remainder Theorem

Question

1)2x^2+6x+3 will have the same remainder when divided by x+p and x-2q. p does not equal to -2q. So what is p-2q?
Answer

Remainder Theorem for you is : if f(x) a polynomial is divided by a linear divisor x-a the remainder is f(a)

So f(x) 2x^2+6x + 3
x+p = x-(-p)
f(-p) = 2(-p)^2 -6p + 3
=2p^2 -6p + 3

f(2q) = 2(2q)^2 + 6.2q + 3
= 8q^2 + 12q + 3

now it seems they have the same remainder so : 2p^2 -6p + 3 =8p^2 +12q +3
2p^2-8q^2 -6p -12q = 0
2(p^2 -4q^) -6(p + 2q) = 0
or 2(p-2q)(p+2q) -6(p + 2q) =0
or (p+2q)(2(p-2q) - 6) =0
so 2(p-2q) -6 =0
or p-2q =6/2
or p-2q =3

Problem#4 Interesting Problem

Here's an interesting problem at Yahoo Answer that I solved this morning. Comments and suggestions and a better approach welcome :)

Substitute 0-9 for the alphabets below using all the digits and repeating none.

A B C
D E F
-----------
sum
-----------
G H I J
SolutionSo easily G is 1 . 0 should be in the Answer becuase if 0 + 2 or something will give result 2. so we will have repeatiton.So lets put H = 0


So
ABC
DEF
--------
10 I J
A + D = 10 .. cant be if we do not have a carry over from the sum of B + E . So A = 4 D = 5.

we are left with 2,3,6,7,8,9

7 + 8 = 15 but if we have + 1 as carry over we can have 16 or 6
left with 2, 3, 9 9 + 3 is 12 and 1 can be used as carry over.

So
473
+589
---------
1062
--------
there can be many combinations
479
583
-----
1062

483
579
-------
and so on.

http://tinyurl.com/6k2vqy8

Problem#3 Absolute value problem

Equation Involving Absolute Value?

Solve |x+|3x-2|| = 2
Solution:

 

|x+|3x-2|| = 2
x^2+2x|3x-2|+9x^2-12x+4= 4
10x^2-12x+2x|3x-2|=0
2x(5x-6+|3x-2|)=0
x=0 or (5x-6+|3x-2|)=0
x=0 or |3x-2|=6-5x
x=0 or 3x-2=6-5x or 2-3x=6-5x
x=0 or x=1 or x=2(rejected)

 

Problem#2 Show that 1/log_6(24)+1/log_12(24)+1/log_824=2

Source: Yahoo Answers. The question and my answer here.

Show that 1/log_6(24)+1/log_12(24)+1/log_824=2

My answer:

Remember:
1/log_A(B)=1/(log_m(B)/log_m(A)
=log_m(A)/log_m(B)

To show that 1/log_6(24)+1/log_12(24)+1/log_824=2
LHS :
1/log_6(24)+1/log_12(24)+1/log_824
=log6/log24+log12/log24+log8/log24
=(log6+log12+log8)/log24

Now remember logA+logB=log(AxB)
Hence
=log(6x12x8)/log24
=log576/log24
=log24^2/log24
=2log24/log24
=2 :)

Source: http://tinyurl.com/6box2u5

Problem#1 What is the remainder when you divide 5(raised to 125) by 124?

Source: Yahoo Answers. The question and my answer here.

What is the remainder when you divide 5(raised to 125) by 124?

Here, 125 is the power of 5. Please tell the method too.

My answer:

5^125 = 5^(3.41 + 2) = 5^3.41 . 5^2
= 125^41.5^2
= (124 + 1)^41 . 5^2
now if you know 124 will come in all the terms except the last which is 1..so
(124^41 + a^124^43 ..... + 1).25
so if you divide this equation by 124 remainder is 25

http://answers.yahoo.com/question/index;_ylt=AvkqN_1zj9KcDbWoYacjikTsy6IX;_ylv=3?qid=20110608034740AAGONcc