Given the polynomial x^2+10x+k
Find all the values of k so that this given polynomial is factorable
we need 2 values which have a sum of 10
how about 6 and 4
x^2 + 10x + 24 = (x+6)(x+4) , k = 24
how about 2 and 8
x^2 + 10x + 16 = (x+2)(x+8) , k = 16
how about -12 and 22
x^2 + 10x - 264 = (x+22)(x-12) , k = -264
etc.
an infinite number of different values of k are possible
To determine the values of k for which the polynomial x^2+10x+k is factorable, we need to find its discriminant. The discriminant of a quadratic equation of the form ax^2+bx+c is given by the formula Δ = b^2 - 4ac.
In this case, the quadratic equation is x^2 + 10x + k, where a = 1, b = 10, and c = k. Therefore, the discriminant Δ can be calculated as follows:
Δ = (10)^2 - 4(1)(k)
= 100 - 4k
= -4k + 100
For the quadratic equation to be factorable, the discriminant Δ must be greater than or equal to zero (Δ ≥ 0). So we have:
-4k + 100 ≥ 0
Now, let's solve this inequality to find the range of k values that satisfy it:
-4k ≥ -100 (subtract 100 from both sides)
k ≤ 25 (divide both sides by -4; note that when dividing an inequality by a negative number, the direction of the inequality flips)
Therefore, the polynomial x^2 + 10x + k is factorable for all values of k that are less than or equal to 25.