x^2 + kx - 19 Find all values of k so that each polynomial can be factored using integers.
First, write down two sets of parentheses to indicate the product.
( )( )
Since the first term in the trinomial is the product of the first terms of the binomials, you enter x as the first
term of each binomial.
(x )(x )
The product of the last terms of the binomials must equal -19.
and their sum must equal -18, and one of the binomials' terms has to be negative. Four different pairs of
factors have a product that equals -19.
Only two integers numbers with the product= -19 and sum= -18 is:
-19 and 1
So:
x^2+kx-19=(x-19)(x+1)=
x*x-19x+x-19= x^2-18x-19
k= -18
wallah ma ba3raf
To find all values of k such that the polynomial x^2 + kx - 19 can be factored using integers, we need to consider the factors of the constant term (-19) and check if they can be combined to give the coefficient of the linear term (k).
The factor pairs of -19 are: (-1, 19) and (1, -19).
Let's try each pair and see if they satisfy the conditions:
1. When the factor pair is (-1, 19):
- We need to find two integers m and n such that m * n = -19 and m + n = -1.
- The only combination that satisfies these conditions is m = -1 and n = 19.
- Therefore, when k = -1, the polynomial x^2 - x - 19 can be factored as (x - 19)(x + 1).
2. When the factor pair is (1, -19):
- We need to find two integers m and n such that m * n = -19 and m + n = 1.
- The only combination that satisfies these conditions is m = -1 and n = 19.
- Therefore, when k = 1, the polynomial x^2 + x - 19 can be factored as (x + 19)(x - 1).
Thus, the values of k that allow the polynomial x^2 + kx - 19 to be factored using integers are k = -1 and k = 1.
To find the values of k such that the polynomial x^2 + kx - 19 can be factored using integers, we need to determine if the polynomial has integer roots.
Let's start by factoring the polynomial x^2 + kx - 19 where the factorization is of the form:
(x + p)(x + q) = x^2 + (p + q)x + pq.
Comparing the coefficients, we have:
- (p + q) = k,
- pq = -19.
Now, we need to find all the combinations of p and q that satisfy the above conditions and check if they are integers.
The factors of -19 are:
1, -1, 19, -19.
So, the possible combinations for (p, q) would be:
(p, q) = (1, -19), (-1, 19), (19, -1), (-19, 1).
Let's check if these combinations satisfy the condition p + q = k and if they give integer values of k:
For (p, q) = (1, -19): p + q = 1 + (-19) = -18. This does not give an integer value for k.
For (p, q) = (-1, 19): p + q = (-1) + 19 = 18. This does not give an integer value for k.
For (p, q) = (19, -1): p + q = 19 + (-1) = 18. This does not give an integer value for k.
For (p, q) = (-19, 1): p + q = (-19) + 1 = -18. This does not give an integer value for k.
Therefore, there are no values of k such that the polynomial x^2 + kx - 19 can be factored using integers.