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-/[]\- Logic chapters 1 to 5 re- appear not in sequence, as is or longer, in Volume 1A, Pattern Based Reason, Bon Appetite. Logic
Mastery Logic mastery makes the hard, easier. Logic mastery leads to better, stronger and richer comprehension. Logic mastery improves reading and writing. Logic mastery ease learning difficulties. Logic mastery gives a headstart. In sum, logic mastery will develops critical thinking, improve reading and writing, and give a firmer base for work and studies at many levels. Good luck. After logic, (a) continue reading Three Skills for Algebra, chapters 8 to 14 and do so alongside site area on solving liinear Equations ; or (b) see this calculus starter lesson and Volume 3, Why Slopes & More Math, chapters 2 to 6;
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-/[]\- What may be learnt and when depends on how skills and concepts are developed. Making the hard easier and clearer will allow earlier & richer development of skills and concepts. Try the Twiddla
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Factoring Quadratics by inspection
Factoring by inspection is based on the backwards or indirect use of the identity (x+A)(x+B) = x2+(A+B)x + AB. that holds for all real numbers x, A and B. Why it holds is indicated as follows. Column Multiplication Method Explanation Geometric Demonstration of (x+A)(x+B) = x2+(A+B)x
+ AB
For x, A and B all positive, the area of the large rectangle is (x+A)(x+B) or the sum of the areas of the small rectangle. This implies (x+A)(x+B) = x2+(A+B)x + AB. The condition that x, A and B all be positive can be removed if one uses the distributive law twice to obtain this result
Factoring x2+bx + c by Inspection
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| p 0,1 |
m 0,1,2 |
n 0,1 |
A = (-1)p3m2n . | B = -18/A | A+B |
| 0 | 0 | 0 | (-1)03020 = 1 | -18 | -17 |
| 1 | (-1)03021 = 2 | -9 | -7 | ||
| 1 | 0 | (-1)03120= 3 | -6 | -3 | |
| 1 | (-1)03121 = 6 | -3 | 3 | ||
| 2 | 0 | (-1)03220 = 9 | -2 | 7 | |
| 1 | (-1)03221 =18 | -1 | 17 | ||
| 1 | 0 | 0 | (-1)13020 = 1 | -18 | -17 |
| 1 | (-1)13021 = 2 | -9 | -7 | ||
| 1 | 0 | (-1)13120= 3 | -6 | -3 | |
| 1 | (-1)13121 = 6 | -3 | 3 | ||
| 2 | 0 | (-1)13220 = 9 | -2 | 7 | |
| 1 | (-1)13221 =18 | -1 | 17 | ||
The above table gives all possible combinations of integers A and B whose product AB = -18. We may draw a tree diagram in place of filling a table.
The A+B value of -3 appears in the third row where (A,B) = (3, -6) So
x2-3x - 18 = (x+A)(x+B) = (x+3)(x-6)
Remark: The above method of finding A and B would not work for quadratics of the form x2+bx -18 where b > 0 and little b does not belong to the set of values for A + B above.
Shortcut: The A+B value of -3 appears in also appears in the third row below the line where (A,B) = (-6, 3). Since A and B have opposite signs and the roles of A and B are interchangeable, and since we need only one solution of the equation A+B = -3, we can adopt the convention that A > B in order to eliminate the solution where the interchanging the values of A and B gives a second solution. That convention A > B here eliminates the need to generate the values (-1)p3m2n where p is odd, or -1 since the requirement A and B have opposite signs and A > B forces A to be positive. If you do several examples, this shortcut will become obvious: practice and numerical experience counts.
A further shortcut: When c is negative, the inspection method for factoring x2+bx + c works by generating all pairs of integers (A,B) for which A > 0 and AB = c and seeing if |b| = A+B for one pair of the generated whole number factors.
Third Example: Try to factor x2-9x + 20 by inspection.
Solution: So if AB = 20 = 5122.(-1)2 with both A and B integers number, A must have the form (-1)p5m2n where the fives-exponent m takes the values 0 or 1, and the twos-exponent n takes the values 0, 1 or 2, and the (-1) exponent p is even or odd, say value 0 or 1
| p | m | n | A = (-1)p5m2n . | B = 20/A | A+B |
| 0 | 0 | 0 | (-1)05020 = 1 | 20 | 21 |
| 1 | (-1)05021 = 2 | 10 | 12 | ||
| 2 | (-1)05022 = 4 | 5 | 9 | ||
| 1 | 0 | (-1)051 20= 5 | 4 | 9 | |
| 1 | (-1)05121 = 10 | 2 | 12 | ||
| 2 | (-1)02251 = 20 | 1 | 21 | ||
| 1 | 0 | 0 | (-1)15020 = -1 | -20 | -21 |
| 1 | (-1)15021 = -2 | -10 | -12 | ||
| 2 | (-1)15022 = -4 | -5 | -9 | ||
| 1 | 0 | (-1)151 20= -5 | -4 | -9 | |
| 1 | (-1)15121 = -10 | -2 | -12 | ||
| 2 | (-1)12251 = -20 | -1 | -21 |
Here the values of A increase in the third column but that does not always occur. Your may draw a tree diagram in place of filling a table. We see that the combination (A,B) = (4, 5) gives A+B = 9 and AB = 20. So the foregoing implies
x2+9x + 20 = (x+A)(x+B) = (x+4)(x+5)
when (A,B) = (4, 5). The same result follows from swapping the values of A and B, that is taking (A,B) = (5,4) instead.
Remark: The above method of finding A and B would not work for quadratics of the form x2+bx + 20 where b > 0 and little b does not belong to the set of values for A + B above.
Shortcut: When c is positive, the inspection method for factoring x2+bx + c works by generating all pairs of whole numbers (A,B) for which AB = c and seeing if |b| = A+B for one pair of the generated whole number factors.
Shortcut for third example: The value of C = 20 is positive. So if A and B are integers with AB = C, we know that AB = C gives (-A)(-B) = C as well, and vice-versa. We also know that A and B will have the same sign. The prime decomposition of 20 = 5122. So if AB = 20 with both A and B being whole number, A must have the form 5m2n where the five exponent m takes the values 0 or 1, and the two exponent n takes the values 0, 1 or 2.
| m | n | A = 5m2n. | B = 20/A | A+B |
| 0 | 0 | 5020 = 1 | 20 | 21 |
| 1 | 5021 = 2 | 10 | 12 | |
| 2 | 5022 = 4 | 5 | 9 | |
| 1 | 0 | 51 20= 5 | 4 | 9 |
| 1 | 5121 = 10 | 2 | 12 | |
| 2 | 2251 = 20 | 1 | 21 |
We see that the coefficient of x in x2-9x + 20 is -9 is not in the table of values for A+B, but its negative +9 is. So (A,B) = (-4, -5) gives A+B = - 9 and AB = 20 as required.
Here want ax2+bx + c = (Ax+B)(Cx+D)
Now want the product of the right side to equal the left hand side.
D +Cx
B+ Ax (times)
BD +BCx
ADx
+ ACx2
BD +(BC +AD)x + ACx2 =a x2+bx + c
So we require BD = c and AC = a. So when a, b and c are integers, there will be a product (Ax+B)(Cx+D) = ax2+bx + c with A, B, C and D integers provided there is a pair (A, C) of complementary factors of a; another pair (B,D) of complementary factors of c such that BC +AD = b.
The foregoing suggests we generate and combine all complementary factors of (A, C) of a with all complementary factors (B, D) of the coefficient c, and see if the coefficient b appears as a value of BC +AD for one of the combinations.
The equality C = a/A and D = c/B implies we need to evaluate the expression
BC +AD = Ba/A + Ac/B
for all integer factors A of a and all integer factors B of c.
Lessons Elsewhere:
Explore this last thought further if you wish. I am going stop here and recommend that you visit visit purplemath lessons on factoring quadratics: the simple case the hard case, the weird case. Multiple views are better than one. The foregoing may be the hard or weird case - not part of mathematics 436. but the simple case is.
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