quick question (1 Viewer)

[5647382910]

this is from a worked examle concerning the factor theorem, i understand the question except the part in red:

question: find one zero, and hence express p(x) = x^3 - x^2 - 14x + 24 as a product of linear factors.
solution: since the coefficient of x^3 is unity, the only possible zeroes involve factors of 24, i.e, 1,2,3,4,6,8,12,24 ............................................. i understand the rest

could someone explain to me wat this means and how do we know it ony involves such factors; also what if the coefficient of x isnt unity?

 

[micuzzo]

umm im a bit rough.. havent done ploys in a while... but what i think your asking is how it says the possible roots come from 24... well its because 24 is the independent term and for such questions the factors of the independent term will be the factors of P(x)


hope it helps

 

[Trebla]

Suppose a, b and c are the roots of the monic cubic equation P(x) = 0
Then P(x) = (x - a)(x - b)(x - c)
Now the constant term (independent of x) is - abc, so this means that a, b and c are all factors of the constant term.
So for a numerical example, if P(x) is monic and has roots 3, 2 and 1, then
P(x) = (x - 1)(x - 2)(x - 3)
= x³ + ............. - 6
Notice that the roots 1, 2 and 3 are all possible factors of - 6.

Now if P(x) is not monic with roots a, b and c
Then P(x) = k(x - a)(x - b)(x - c) for some constant k (note the leading co-efficient is k)
Now the constant term is - kabc, so this means a, b and c are factors of the constant term, but to find them you have to remove the k.
Using a numerical example, if P(x) has a leading co-efficient of 2 and has roots 3, 2 and 1, then
P(x) = 2(x - 1)(x - 2)(x - 3)
= 2x³ + ............. - 12
Notice that the roots 1, 2 and 3 are all possible factors of - 12. However, to find them you need to get rid of that 2, which simplifies the problem to the monic case since to find factors you solve
P(x) = 2[x³ + ............. - 6] = 0
=> x³ + ............. - 6 = 0 which is the same as the monic case so you need to find factors of - 6 rather than - 12 which simplifies the problem.

So in your example: P(x) = x³ - x² - 14x + 24
Test x = 2:
P(2) = 8 - 4 - 28 + 24
= 0
Hence (x - 2) is a factor.
Now suppose P(x) = 2x³ - x² - 14x + 24 instead...then we can simplify the problem to 2[x³ - (1/2)x² - 7x + 12] in which case we have to test the factors of 12.

 

[5647382910]

Suppose a, b and c are the roots of the monic cubic equation P(x) = 0
Then P(x) = (x - a)(x - b)(x - c)
Now the constant term (independent of x) is - abc, so this means that a, b and c are all factors of the constant term.
So for a numerical example, if P(x) is monic and has roots 3, 2 and 1, then
P(x) = (x - 1)(x - 2)(x - 3)
= x³ + ............. - 6
Notice that the roots 1, 2 and 3 are all possible factors of - 6.

Now if P(x) is not monic with roots a, b and c
Then P(x) = k(x - a)(x - b)(x - c) for some constant k (note the leading co-efficient is k)
Now the constant term is - kabc, so this means a, b and c are factors of the constant term, but to find them you have to remove the k.
Using a numerical example, if P(x) has a leading co-efficient of 2 and has roots 3, 2 and 1, then
P(x) = 2(x - 1)(x - 2)(x - 3)
= 2x³ + ............. - 12
Notice that the roots 1, 2 and 3 are all possible factors of - 12. However, to find them you need to get rid of that 2, which simplifies the problem to the monic case since to find factors you solve
P(x) = 2[x³ + ............. - 6] = 0
=> x³ + ............. - 6 = 0 which is the same as the monic case so you need to find factors of - 6 rather than - 12 which simplifies the problem.

So in your example: P(x) = x³ - x² - 14x + 24
Test x = 2:
P(2) = 8 - 4 - 28 + 24
= 0
Hence (x - 2) is a factor.
Now suppose P(x) = 2x³ - x² - 14x + 24 instead...then we can simplify the problem to 2[x³ - (1/2)x² - 7x + 12] in which case we have to test the factors of 12.

you have cleared it up 110% for me, ur help is appreciated greatly