what value(s) of k make the expression below factorable?There are five correct answers. kx2+10x+1

Question

The problem is for a extra credit math problem. And having a hard time on it.

Answer 1

Since kx2+10x+1 is factorable, let's say the factors are (ax+1) and (bx+1)

 

[comment: you will need +1 in each factor to produce the +1 in kx2+10x+1]

 

Now, since (ax+1) and (bx+1) are factors of kx2+10x+1

(ax+1)(bx+1) = kx2+10x+1

abx2 + (a+b)x + 1 = kx2+10x+1

Equating similar terms,

ab = k                      eq1

a+b = 10                  eq2

Substitute value of b from eq2 into eq1

a(10-a) = k

Now, think about the values of a, b and k

First of all, k can not be zero, otherwise your quadratic would reduce to a linear equation.

If we assume, that k is a natural number, then

when a=1, b=9, k=9

when a=2, b=8, k=16

when a=3, b=7, k=21

when a=4, b=6, k=24

when a=5, b=5, k=25

These are your five solutions.

 

What happens, when a is 6, 7, ...

when a=6, b=4, k=24, so you get the same solution as when a =4 and b=6

an so on...