■identify the values that should be written to complete the x diagram. on the top: –28–31 on the bottom: –28–31 on the sides: –14 and 2–7 and 4–4 and 7 ■rewrite the expression using the numbers on the sides of the x diagram. ■use double grouping to factor the four terms. x2 – 3x – 28 = (x – 14)(x + 2)(x – 7)(x + 4)(x + 7)(x – 4)

The given quadratic is ...

... x² -3x -28 . . . . . a=1, b=-3, c=-28

a) The value at the top of the X diagram is the product a·c = 1·(-28) = -28.

The value at the bottom of the X diagram is the coefficient b = -3.

The values on the sides of the diagram are the factors of -28 that add up to make -3. These are -7 and 4. That is, ...

b) Since the two values on the sides of the diagram add up to give "b", the value of "b" in the equation can be rewritten as the sum of these two numbers. Doing that, we have ...

... x² -3x -28

... = x² +(-7+4)x -28

... = x² -7x +4x -28 . . . . . . order does not matter. It could also be x² +4x -7x -28

c) We can group pairs of terms in the rewritten expression and factor each pair.

... = (x² -7x) +(4x -28) . . . . . first pair has a common factor of x; second pair, 4

... = x(x -7) +4(x -7) . . . . . . . these terms now have a common factor: (x -7)

... = (x +4)(x -7) . . . . . . . . . . the complete factorization of x² -3x -28

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Comment on extended application

When a ≠ 1, there are a couple of additional steps. Effectively, you reduce the fractions ax/p and ax/q to a'x/p' and a''x/q'', then rewrite those as the factors (a'x+p')(a''x+q'') where the ' and '' numbers are after reducing the fractions. (The numbers may be the same as the original if the fraction cannot be reduced.) As a check, a'·a'' = a, and p'·q'' = c.

its a straight line parallel to the time axis

inverse operation of addition