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 May 15th, 2017, 10:17 AM #11 Member     Joined: Aug 2016 From: afghanistan Posts: 53 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: 6,667 Thanks: 2170 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: 53 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: 507 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: 6,667 Thanks: 2170 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: 1,957 Thanks: 639 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.

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