What are 2 values of b that will make 2x^2 - bx - 20 factorable? Explain your answer.
A: ?
(2x-20)(x+1) for example would be
2 x^2 - 18 x - 20
(2 x - 10)(x + 2) would be
2 x^2 - 6 x - 20
so 18 or 6 would work
12
To determine the values of b that will make the quadratic polynomial 2x^2 - bx - 20 factorable, we can use the fact that if a quadratic polynomial is factorable, its discriminant must be a perfect square.
The discriminant of a quadratic polynomial in the form ax^2 + bx + c is given by the expression b^2 - 4ac. In this case, the polynomial is 2x^2 - bx - 20, so the discriminant is b^2 - 4(2)(-20) = b^2 + 160.
For the discriminant to be a perfect square, there must exist an integer d such that d^2 = b^2 + 160. Rearranging this equation, we have b^2 = d^2 - 160.
We can now look for pairs of values for b and d that satisfy this equation. One possible approach is to find perfect squares that are 160 less than another perfect square. We can start by listing some perfect squares greater than 160:
11^2 = 121
12^2 = 144
13^2 = 169
14^2 = 196
Next, we subtract 160 from each of these perfect squares:
121 - 160 = -39
144 - 160 = -16 (not a perfect square)
169 - 160 = 9
196 - 160 = 36 (not a perfect square)
We can see that when d = 13, (13^2 = 169), b^2 = d^2 - 160 = 169 - 160 = 9. Therefore, one value of b that will make the polynomial factorable is b = 3, because 3^2 = 9.
Similarly, when d = 11, (11^2 = 121), b^2 = d^2 - 160 = 121 - 160 = -39. Since the square of a real number cannot be negative, there are no real roots for b in this case.
In summary, the two values of b that will make the polynomial 2x^2 - bx - 20 factorable are b = 3 and b = -3.