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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 May 15th, 2017, 10:17 AM #11 Member     Joined: Aug 2016 From: afghanistan Posts: 55 Thanks: 1 I think i got it from the same set of text books , Factorise 6x2 + 17x + 5 by splitting the middle term (By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17. So, 6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5 = 6x2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5) I guess that,s it ?
 May 15th, 2017, 11:32 AM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,775 Thanks: 2194 Math Focus: Mainly analysis and algebra Looks ok to me, but it's not what I'd call the most accessible technique. Try this: \begin{align*} 6x^2 + 17x + 5 &= \frac66(6x^2 + 17x + 5) &(\text{\frac66 because 6 is the coefficient of x^2}) \\ &= \frac16 \big((6x)^2 + 17(6x) + 30\big) &\text{now write u=6x} \\ &= \frac16(u^2 + 17x + 30) &\text{and factor the quadratic in parentheses} \\ &= \frac16(u+2)(u+15) &\text{replace the 6xs} \\ &= \frac16(6x+2)(6x+15) \\ &= \frac12(6x+2)\frac13(6x+15) \\ &= (3x+1)(2x+5) \end{align*} Thanks from porchgirl
 May 15th, 2017, 03:02 PM #13 Member     Joined: Aug 2016 From: afghanistan Posts: 55 Thanks: 1 Thanks , I am going to re learn math from those set of textbooks and if i have more doubts i will come back here and ask
 May 21st, 2017, 12:38 PM #14 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 522 Thanks: 74 In the first post what do the letters on the right like "AT" and "DL" mean?
 May 21st, 2017, 01:38 PM #15 Math Team   Joined: Dec 2013 From: Colombia Posts: 6,775 Thanks: 2194 Math Focus: Mainly analysis and algebra AT is something to do with addition. DL: distributivity law. ALA: associativity law of addition. Etc..
 May 22nd, 2017, 03:36 AM #16 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 1,987 Thanks: 646 Math Focus: Physics, mathematical modelling, numerical and computational solutions The method I was taught at school was to make little tables of pairs of factors of the first and third coefficients in the quadratic. This isn't any different from the splitting method, but it provides a means of recording the various options for multiplying the factors together so you can eventually find the option which yields the second coefficient. The method is rather long-winded, but it helps a lot when you can't figure out the factorisation quickly in your head or you have a horrible quadratic. So for $\displaystyle 6x^2 + 17x + 5$ you'd draw a table like so: $\displaystyle \begin{array}{l|ll} 3 & 5 & 1 \\ \hline 2 & 1 & 5 \end{array}$ and $\displaystyle \begin{array}{l|ll} 6 & 5 & 1 \\ \hline 1 & 1 & 5 \end{array}$ where, in the first table, the 3 and 2 on the left-hand side are factors of 6 (the first coefficient) and the 1 and 5 are the factors of 5 (the third coefficient). The second table is the same but for the other combination of factors for the 6 (1 and 6). For each set of factors of the third coefficient, you need to consider two possible combinations, hence the two entries in both tables for the 1 and 5. The first thing you check is the sign of the third coefficient. In our case it is a +. This means that in the next step we must add to obtain the second coefficient. If it were negative, we would have to subtract instead. Then, you test the various cases to see if you can obtain the second coefficient (17 in this case). Let's start with the first table... You read across the table column by column and multiply diagonally to check the values obtained by the cross multiplication of the brackets... so in our example case, we look at the first column in the table following the 3 and 2 and multiply diagonally to get $\displaystyle 3 \times 1 = 3$ and $\displaystyle 2 \times 5 = 10$ Can we add 3 and 10 to give the second coefficient (17)? The answer is no, so you discard that column (cross it out or put a cross above it or something) and move onto the next column. Then you get $\displaystyle 3 \times 5 = 15$ and $\displaystyle 2 \times 1 = 2$ Can we add 15 and 2 to give the second coefficient (17)? The answer is yes, so you then form your brackets by reading off from the table: $\displaystyle 6x^2 + 17x + 5 = (3x+1)(2x+5)$ You'd also worry about sign placement here, but it's easy to figure out once you have the values you're after.
 May 23rd, 2017, 12:47 AM #17 Member     Joined: Aug 2016 From: afghanistan Posts: 55 Thanks: 1 Thanks for the replies , The problem is i haven't found one book that explains all these things in one book . I am trying to learn from many different books and websites .Still trying to narrow it down , so that i can look at one place and try to learn . Can i put everything together one time ? I started from here , http://ncert.nic.in/textbook/textbook.htm National Council Of Educational Research And Training :: Home SOLVING EQUATIONS ’To factor’ means to break up into multiples. Factors of natural numbers The numbers other than 1 whose only factors are 1 and the number itself are called Prime numbers Numbers having more than two factors are called Composite numbers. Greatest common factor The Greatest Common Factor (GCF) of two or more given numbers is the greatest of their common factors Lowest Common Multiple The Lowest Common Multiple (LCM) of two or more given numbers is the lowest (or smallest or least) of their common multiples. You will remember what you learnt about factors in Class VI. Let us take a natural number, say 30, and write it as a product of other natural numbers, say 30 = 2 × 15 = 3 × 10 = 5 × 6 Thus, 1, 2, 3, 5, 6, 10, 15 and 30 are the factors of 30. Of these, 2, 3 and 5 are the prime factors of 30 (Why?) A number written as a product of prime factors is said to be in the prime factor form; for example, 30 written as 2 × 3 × 5 is in the prime factor form. The prime factor form of 70 is 2 × 5 × 7. The prime factor form of 90 is 2 × 3 × 3 × 5, and so on. Similarly, we can express algebraic expressions as products of their factors. This is what we shall learn to do in this chapter. Factors of algebraic expressions We have seen in Class VII that in algebraic expressions, terms are formed as products of factors. For example, in the algebraic expression 5xy + 3x the term 5xy has been formed by the factors 5, x and y, i.e., 5xy = 5 * x * y Observe that the factors 5, x and y of 5xy cannot further be expressed as a product of factors. We may say that 5, x and y are ‘prime’ factors of 5xy. In algebraic expressions, we use the word ‘irreducible’ in place of ‘prime’. We say that 5 × x × y is the irreducible form of 5xy. Note 5 × (xy) is not an irreducible form of 5xy, since the factor xy can be further expressed as a product of x and y, i.e., xy = x × y. What is Factorisation? When we factorise an algebraic expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions. Expressions like 3xy, 5x2y , 2x (y + 2), 5 (y + 1) (x + 2) are already in factor form. Their factors can be just read off from them, as we already know. On the other hand consider expressions like 2x + 4, 3x + 3y, x2 + 5x, x2 + 5x + 6. It is not obvious what their factors are. We need to develop systematic methods to factorise these expressions, i.e., to find their factors. Methods of Factoring Method of common factors Factorisation by regrouping terms Factorisation using identities Factors of the form ( x + a) ( x + b) Factor by Splitting Factorise 6x2 + 17x + 5 by splitting the middle term (By splitting method) : If we can find two numbers p and q such that p + q = 17 and pq = 6 × 5 = 30, then we can get the factors So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5 and 6. Of these pairs, 2 and 15 will give us p + q = 17. So, 6x2 + 17x + 5 = 6x2 + (2 + 15)x + 5 = 6x2 + 2x + 15x + 5 = 2x(3x + 1) + 5(3x + 1) = (3x + 1) (2x + 5) Practicing like that should help , right ?
May 23rd, 2017, 06:56 AM   #18
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 Originally Posted by porchgirl Thanks for the replies , The problem is i haven't found one book that explains all these things in one book . I am trying to learn from many different books and websites .Still trying to narrow it down , so that i can look at one place and try to learn .
I can't help thinking that you are trying to memorize everything as a collection of unrelated facts, when the point is to develop your mathematical sense to the point that if you didn't remember a formula, you could figure it out on your own.

I don't know how to tell you how to do this, but in my opinion you should be stepping back and trying to gain some intuition, not just looking for a huge book full of rules and formulas.

 May 23rd, 2017, 07:33 AM #19 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 1,987 Thanks: 646 Math Focus: Physics, mathematical modelling, numerical and computational solutions I think the reason that you haven't found one book that explains everything is because you're looking at a mixture of simpler content (factors and multiples, factorising linear equations and quadratics) and much more complex content (the cubic polynomials, the Pascal polynomials, etc). All the information you posted is (more or less) a complete look at factorisation, at least as far as I know about it. The only thing which I'd add to your list is "completing the square", which is not factorisation but is nevertheless a very, very useful method to learn. When I was at school I learned how to complete the square at about the same time as how to factorise quadratics. As for factorising quadratics, if you're comfortable with the splitting method, stick with that and ignore my previous post with the tables. Lastly.... do lots of questions! It helps a lot.

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