Can someone please help me with this question?

Question

Find all values of 'k' so that the trinomial x^2+kx-35 can be factored using integers.

Answer 1

This is an interesting question.
Let's look at the factors of 35, which are 5 and 7 and 1 and 35

So, we're looking at (x-5)*(x+7) as one possibility (note the negative sign in front of the 35. This expands to x^2 -5x+7x - 35, or
x^2 + 2x - 35.
So k=2 is ONE possibility.
Another arises from (x+5)*(x-7) , where k=-2
(x-1)*(x+35) I'll let you figure out k
and (x+1)*(x-35), the fourth value of k

Answer 2

Basically, what you want is the discriminant, b^2 - 4ac, to be a perfect square. Note that this only guarantees that the roots are RATIONAL and not necessarily integers, but it does eliminate the irrational roots.

For ax^2 + bx + c = 0, a = 1, b = k, and c = -35. Thus,

b^2 - 4ac = k^2 - 4(1)(-35) = k^2 + 140

So we need integer solutions k and z for

k^2 + 140 = z^2

Swapping terms,

k^2 - z^2 = -140

Multiplying -1 to both sides,

z^2 - k^2 = 140

EDIT: SOLVED!!

Therefore,

(z - k) (z + k) = 140

This immediately tells us that (z - k) (z + k) can represents factors of 140.

Since z and k both have to be integers, it follows that

z - k = f1, and
z + k = f2

Where f1 and f2 are factors of 140. if we solve this system of equations for z and k, we obtain

z = [f2 + f1]/2
k = [f2 - f1]/2

At this point, we can automatically see that the sum and difference of the factors MUST be an even number in order for us to get integer solutions.

The prime factorization of 140 is 2 x 2 x 5 x 7, so let's exhaust our possibilities of what two integers make 140. We can automatically eliminate 1 x 140 and 140 x 1, since those factors add up to an odd number, which gets rejected as per above. Remember that, by inclusion, if (f1 x f2) gets rejected, so does (f2 x f1), which means I don't have to list all permutations of the factors.

140 = 2 x 2 x 5 x 7, therefore, our pairs are:
2 x 70, 70 x 2
5 x 28 (reject)
7 x 20 (reject)
4 x 35 (reject)
10 x 14, 14 x 10

So our only working factors are 2 x 70, 70 x 2, 10 x 14, 14 x 10. Recall that

z = [f2 + f1]/2
k = [f2 - f1]/2

Plugging in f1 = 2 and f2 = 70, we obtain
k = (70 - 2) / 2 = 68/2 = 34

f1 = 70, f2 = 2:
k = (2 - 70) / 2 = -68/2 = -34

f1 = 10 and f2 = 14
k = (14 - 10) / 2 = 4/2 = 2

f1 = 14 and f2 = 10
k = (10 - 14) / 2 = -4/2 = -2

Therefore, our integer solutions for k are

k = 34, -34, 2, -2

And our factorable trinomials are

x^2 + 34x - 35
x^2 - 34x - 35
x^2 + 2x - 35
x^2 - 2x - 35

Answer 3

The equation can be re-written as follows"

x^2 + kx - 35 = (x+k1)(x+k2)

From this we can see that k = k1 + k2 and k1*k2 = -35.

There are four pairs for k1 and k2 that satisfy the last relationship and are integers -- (1) 5 and -7, (ii) -5 and 7, (iii) 1 and -35, and (iv) -1 and 35.

Therefore, k can take on the following values:

2 (=5-7)
-2 (=7-5)
-34 (=1-35)
34 (=35-1)

Answer 4

Let x^2+kx-35 = (x-m)(x-n) = x^2 - (m+n)x + mn.
As long as both mn = -35 and (m and n) both are an integer, then you get k.

k = ±[2, 34]

--------------------------------
Puggy,

That the discriminant is a perfect square cannot guarantee you can factor the trinomial into integers.

Answer 5

K must be a difference of the interger factors of 35- that is, 5 - 7 or 7 - 5. Answer is ±2.

Answer 6

x^2+kx-35
You have to split -35 in factors:
-35 = -5 x +7 = +5 x -7 = -1 x35 = +1 x -35
When you add the two factors, you find possible k:
+2 , -2 , +34 , -34

Th

Answer 7

We need all the different combinations that give us -35 as a product.

k is the sum

-1 * 35, k=34, (x-1)(x+35)=x^2+34x-35
-5 * 7, k=2, (x-5)(x+7)=x^2+2x-35
5 * -7, k=-2, (x+5)(x-7)=x^2-2x-35
1 * -35, k=-34, (x+1)(x-35)=x^2-34x-35

Answer 8

x^2+kx-35

factors of -35 are
1*-35 (x+1)(x-35)=x^2-34x-35 k=-34
-1*35 (x-1)(x+35)=x^2+34x-35 k=34
5*-7 (x+5)(x-7)=x^2-2x-35 k=-2
-5*7 (x-5)(x+7)=x^2+2x-35 k=2

Answer 9

x^2+kx-35=(x+5 )(x- 7)
then k=-2
or x^2+kx-35=(x+7)(x-5)
then k= +2
all values of (k) are [2,-2]