How To Factor 2x^3-5x^2-125

How do you factorise 3x^2 + 5x+2?

There is a method so you can obtain (x-1) (3x+2)3x^2 +5x +2Take the first coefficient a, and the last coefficient c,a=3 and c=2Times them together =6Now find which factors of 6 add up to coefficient b. b=52x3=6 and 2+3=5Now split up the equation:3x^2 +3x +2x +2 =0Factorise the first two terms, 3x^2 +3x3x(x+1)Then, just write down the bracket again next to it.3x (x+1) ___ (x+1)Fill in the gap, what times x+1 is 2x+2?3x(x+1) +2(x+1)Now finally, write the bracket (x+1) and put whatever is left in another bracket.(X+1) (3x+2)This method works for most quadratics with a coefficient a which is more than 1.

Five congruent rectangles are drawn inside a big rectangle of perimeter 165. what will be the perimeter of one of the five rectangle?

There are several possible solutions,The perimeter of the original rectangle is 165. So the combined length and the width add up to 165/2 or 82.5.One rectangle can be 42.5 x 40. The 5 congruent smaller rectangles will be 42.5x8. The perimeter of each will be (42.5+8)*2 = 101.The next rectangle can be 47.5 x 35. The 5 congruent smaller rectangles will be 47.5x7. The perimeter of each will be (47.5+7)*2 = 109.The next rectangle can be 52.5 x 30. The 5 congruent smaller rectangles will be 52.5x6. The perimeter of each will be (52.5+6)*2 = 117.The next rectangle can be 57.5 x 25. The 5 congruent smaller rectangles will be 57.5x5. The perimeter of each will be (57.5+5)*2 = 125.The next rectangle can be 62.5 x 20. The 5 congruent smaller rectangles will be 62.5x4. The perimeter of each will be (62.5+4)*2 = 133.The next rectangle can be 67.5 x 15. The 5 congruent smaller rectangles will be 67.5x3. The perimeter of each will be (67.5+3)*2 = 141.The next rectangle can be 72.5 x 10. The 5 congruent smaller rectangles will be 72.5x2. The perimeter of each will be (72.5+2)*2 = 149.The next rectangle can be 52.5 x 30. The 5 congruent smaller rectangles will be 77.5x1. The perimeter of each will be (77.5+1)*2 = 157.

What will be the remainder when [math]2^{31}[/math] is divided by [math]5[/math]?

In very Layman terms,2^1=2, 2^2=4,2^3=8,2^4=162^5=32,2^6=64notice the last digits, the sequence is repeating periodically({1,4,8,6}) at intervals of 4, i.e. the last digit of 2^1 and 2^5 and 2^9 are the same.So now as 31 can be written as 4*7+3, the last digit of 2^31 will be 8. divide 8 by 5 you get a remainder of 3.(any integer with last digit 8 divided by 5 will give a remainder of 3).For generalizing things read up on congruence classes.

What's the remainder when [math]2^{99}[/math] is divided by 33?

Many answers above have explained “how” to solve it. What we also need to understand is “why” it is solved this way. We use the binomial theorem to solve such questions. Let me explain:Let me start with something you already know.(a + b)^2 = a^2 + 2ab + b^2(a + b)^3 = a^3 + 3ba^2 + 3ab^2 + b^3What about (a + b)^4 or (a + b)^5 or higher powers? Binomial theorem just tells us how to expand these expressions. It gives you a general formula:(a + b)^n = a^n + n*a^(n-1)*b + n(n-1)/2*a^(n-2)*b^2 +……..+ n*a*b^(n-1) + b^nI know the above looks intimidating but our concern is limited to the last term of the expression. Notice that every term above is divisible by ‘a’ except for the last term b^n. Every term but the last has ‘a’ as a factor. That is all you need to understand about Binomial Theorem.Now for a quick example before we look at our question:What is the last term when you expand (8 + 1)^20? It is 1^20 (which is just ‘1’).When you expand (8 + 1)^20, is every term divisible by 8? Yes, except for the last term, 1, because every term has 8 as a factor except for the last term.If I divide (8 + 1)^20 by 8, what will be the remainder? Since every term (except for the last one) in the expansion of (8 + 1)^20 is divisible by 8, I can say that (8 + 1)^20 is 1 more than a multiple of 8. Hence the remainder when I divide it by 8 will be 1.Or I can say that when I divide 9^20 (which is just (8 + 1)^20) by 8, the remainder is 1.Now on to the original question: 2^99 has to be divided by 33.We know that 2^5 = 32So 2^99 = 2^4 * 2^95= 16 * 32^19= 16 * (33 - 1)^19Here comes our Binomial theorem. Every term of this expansion will be divisible by 33 except the last term (-1)^19 which is nothing but -1.= 16 [33a + 33b + 33c + … + (- 1)]= 16*33a + 16*33b + 16*33c + … - 16Remainder will be -16.Now, we come to the concept of negative remainders.A negative remainder of -R is the same as a positive remainder of (Divisor - R). So the negative remainder -16 is the same as a positive remainder of 33 - 16 = 17.Here are two posts from my Veritas blog Quarter Wit Quarter Wisdom which discuss Binomial theorem and Negative Remainders:Quarter Wit, Quarter Wisdom: A Remainders Post for the Geek in You!All About Negative Remainders on the GMATShout-out if something is unclear!

If 25x = 125, then what is the value of x, and how?

25x = 125meaning: there are “x" times 25 which makes or equals 125.To know the value of x, we have to subtract or take out 25 from 125 until 125 become 0 , and the number of times 25 is taken out is the value of x.Or add 25’s until 125 is found. The times you add 25 to get 125 is the value of x.Using substraction:First time : take out first “25” from 125;125–25= 100.Second time: take out another 25 from the previous result.100–25= 75.Third time: take out another 25 from the previous result.75–25=50Fourth time: take out another 25 from the previous result i.e. from 50.50–25=25.Fifth time: take out another 25 from the previous result.25–25=0So from the above five time subtraction,we Can see that five “25” adds up together to make 125 or there are five “25” in 125.Therefore,x in 25x=125 is 5.Or simply divide both side by 2525x = 125x/25=125/25x=5