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Pie
3 years ago
7

Find two values for k such that the trinomial x2 – 7x + k can be factored over the integers. Explain your reasoning.

Mathematics
1 answer:
vova2212 [387]3 years ago
6 0

Answer:

1) For a = 1: b = 6 and k = 6, 2) For a = 3: b = 4 and k = 12

Step-by-step explanation:

The polynomial y = x^{2} - 7\cdot x + k is a second-order polynomial of the form (x-a)\cdot (x-b) = x^{2}-(a+b)\cdot x + a\cdot b. By direct comparison, we construct the following system of equations:

a + b = 7 (1)

a\cdot b = k (2)

By (1) we know that there are a family of pairs such that the system of equations is satisfied. Let suppose that both a and b are integers. We assume two arbitrary integers for a:

1) a = 1

b = 7 - a

b = 6

a\cdot b = k

k = (6)\cdot (1)

k = 6

2) a = 3

b = 7 - a

b = 4

a\cdot b = k

k = (3)\cdot (4)

k = 12

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\bf \cfrac{x^2-x-42}{x-6}\implies \cfrac{(x+7)\underline{(x-6)}}{\underline{x-6}}\implies x+7

so.. the left-hand-side does indeed simplify to x+7, so the equation does check out.

however, notice something, for the equation of x+7, when x = 6, we get (6) + 7 which is 13.

BUT for the rational, we get    \bf \cfrac{x^2-x-42}{x-6}\qquad \boxed{x=6}\implies \cfrac{x^2-x-42}{\boxed{6}-6}\implies \stackrel{und efined}{\cfrac{x^2-x-42}{0}}

so, even though the siimplification is correct, the rational or original expression is constrained in its domain.
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