if a rectangular paintings area is given by the polynomial x^2-2x-80 what are the possible dimensions of the painting.
well,
x^2-2x-80 = (x-10)(x+8)
so, ...
why would you set either factor to zero? As long as x > 10 you can satisfy the conditions of the problem.
To determine the possible dimensions of the rectangular painting, we need to factor the given polynomial, which represents the area of the painting.
The given polynomial is x^2 - 2x - 80.
First, we'll factorize the polynomial by finding the values of x that make the polynomial equal to zero. This can be done by factoring or by using the quadratic formula.
Using the quadratic formula, which states that for a quadratic equation in the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / 2a.
In our case, a = 1, b = -2, and c = -80.
Plugging these values into the quadratic formula, we get:
x = (2 ± √((-2)^2 - 4(1)(-80))) / 2(1)
x = (2 ± √(4 + 320)) / 2
x = (2 ± √324) / 2
x = (2 ± 18) / 2.
This simplifies to:
x = (2 + 18) / 2 or x = (2 - 18) / 2,
x = 20/2 or x = -16/2,
x = 10 or x = -8.
So, the possible dimensions of the painting, in terms of length and width, are 10 units by -8 units or -8 units by 10 units. However, dimensions cannot be negative, so the actual possible dimensions for the rectangular painting are 10 units by 8 units.
Solve by factoring out the polynomial, and you would get:
(x+8)(x-10). Set those equal to 0, and x will equal 10 and -8.
Thus, the answer is that the possible dimensions 10 by 8 (you can't have negative dimensions, so change -8 to positive).
Hope this helps!