# Really Important Factoring Question?

The expression 3x^2 - 11x + k can be factored into two linear polynomials with integer coefficients. Determine the possible values of k.

Can someone do this step by step please.

### 2 Answers

- 1 decade agoFavorite Answer
Hi, first of all notice that the given polynomial can be factored as a product of linear factors with integer coefficients if and only if it has rational roots a/b and c/d such that a/b + c/d = 11/3 and (a/b)(c/d) = k/3 where either a = b = 1 or one of them is 3 and the other is 1 (this comes from the fact that the polynomial factors in 3(x - a/b)(x - c/d) and 3 is prime). Assuming a = b = 1 you get the system

a + c = 11/3

ac = k/3

which has clearly no integer solutions. Then you must have for example b = 3 and d = 1 (the choice will not affect the result) and in this case the system becomes

a/3 + c = 11/3

ac = k,

that is

a + 3c = 11

ac = k

with a and c integers. From the first one you get a = 11 - 3c, which substituted in the second one gives you

k = c(11 - 3c),

which as c varies in Z should give you all the possible values of k for which 3x^2 - 11x + k can be factored into linear polynomials with integer coefficients.

I hope this helped you. Bye!

@Roger The Mole: for any integer c if k = c(11 - 3c) the discriminant is

36c^2 - 132c + 121 = (6c - 11)^2,

that is a perfect square. So for example k = -70 (you get it or c = 10) is a good value, which leads to the decomposition

(3x + 10)(x - 7).

Moreover, the polynomial p(x) = x is linear, it's not suspicious, it suits the definition.

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- Roger the MoleLv 71 decade ago
For the expression to be factorable in integers, there must be integer solutions for

3x^2 - 11x + k = 0.

Since this is a quadratic equation, that means that the discriminant must be a perfect square. That is, (121 - 12k ) must be a perfect square.

That is true for k = -4, 0 , 6, 8, and 10.

The solution k = 0 is suspicious, because x = 0 might not be considered one of "two linear polynomials".

The corresponding factorizations, including the case k = 0, are:

(x - 4) (3x + 1)

x (3x - 11)

(x - 3) (3x - 2)

(x - 1) (3x - 8)

(x - 2) (3x - 5)

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