Which of the following quartic functions has x = -2 and x = -3 as its only two real zeros?
a) x^4-5x^3+7x^2-5x+6
b) x^4-5x^3+7x^2+5x-6
c)x^4+5x^3+7x^2+5x+6
d)x^4+5x^3+7x^2-5x-6
The correct answer is b) x^4-5x^3+7x^2+5x-6.
To determine the real zeros of a quartic function, we need to find the values of x that make the function equal to zero. In this case, we are given that x = -2 and x = -3 are the only two real zeros.
By substituting x = -2 and x = -3 into each of the quartic functions, we can determine which function satisfies this condition.
For option a) x^4-5x^3+7x^2-5x+6:
(-2)^4 - 5(-2)^3 + 7(-2)^2 - 5(-2) + 6 = 16 + 40 + 28 + 10 + 6 = 100
(-3)^4 - 5(-3)^3 + 7(-3)^2 - 5(-3) + 6 = 81 + 135 + 63 + 15 + 6 = 300
For option b) x^4-5x^3+7x^2+5x-6:
(-2)^4 - 5(-2)^3 + 7(-2)^2 + 5(-2) - 6 = 16 + 40 + 28 - 10 - 6 = 68
(-3)^4 - 5(-3)^3 + 7(-3)^2 + 5(-3) - 6 = 81 + 135 + 63 - 15 - 6 = 258
For option c) x^4+5x^3+7x^2+5x+6:
(-2)^4 + 5(-2)^3 + 7(-2)^2 + 5(-2) + 6 = 16 - 40 + 28 - 10 + 6 = 0
(-3)^4 + 5(-3)^3 + 7(-3)^2 + 5(-3) + 6 = 81 - 135 + 63 - 15 + 6 = 0
For option d) x^4+5x^3+7x^2-5x-6:
(-2)^4 + 5(-2)^3 + 7(-2)^2 - 5(-2) - 6 = 16 - 40 + 28 + 10 - 6 = 8
(-3)^4 + 5(-3)^3 + 7(-3)^2 - 5(-3) - 6 = 81 - 135 + 63 + 15 - 6 = 18
Out of all the options, only b) x^4-5x^3+7x^2+5x-6 satisfies the condition of having x = -2 and x = -3 as its only two real zeros.