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Note: If a polynomial function passes through the x-axis, the points of intersection are the "real" roots. Real roots can be either rational or irrational. In this lesson, we are interested in finding *rational* roots. 1. Write a fraction bar. 2. List all factors of the constant termin the numerator3. List all factors of the leading4. List all combinations of quotients.These are the possible rational roots coefficient in the denominator The rational root theorem gives us a list of all of the*possible* rational roots of a polynomial. Given the polynomial function: y = 5x3-7x2-5x+7 {1, -1, 7, -7, 1/5, -1/5, 7/5, -7/5} ±1, ±7 ±1, ±5 Write an expression that represents all possible rationalroots of the polynomial: f(x) = 2x3 - 5x2 + x - 6 Factors of -6 ? Factors of 2 ? Write an expression that represents all possible rationalroots of the polynomial: f(x) = 2x3 - 5x2 + x - 6 Factors of -6 Factors of 2 ± ± 1, 2, 3, 6 ? 1, 2 ? There are three ways. How do we know which numbers on the list If the value of the function is zero using any of these threemethods, then the number is a root of the function! 1. Plug In/ Evaluate the function. 2. Divide using synthetic division 3. Graph the related function are actually answers/roots/solutions? Using the function f(x) = x3 - 2x2 - 1x + 2 ± ± 1, 2 ? 1 ? let's find the list of possible rational roots and see which ones actually work Possible rational roots: {1, -1, 2, -2} ? Using the function f(x) = x3 - 2x2 - 1x + 2 Possible rational roots: Find f(1), f(-1), f(2), and f(-2). First find f(1). f(1) = (1)3 - 2(1)2 - 1(1) + 2 f(1) = 1 - 2 - 1 + 2 f(1) = {1, -1, 2, -2} 1 is not a root of the function 1 is a root of the function Select the true statement. Using the function f(x) = x3 - 2x2 - 1x + 2 Possible rational roots: Find f(1), f(-1), f(2), and f(-2). Next find f(-1). f(-1) = (-1)3 - 2(-1)2 - 1(-1) + 2 f(-1) = -1 - 2 + 1 + 2 f(-1) = {1, -1, 2, -2} ? ✅ -1 is not a root of the function -1 is a root of the function Select the true statement. Using the function f(x) = x3 - 2x2 - 1x + 2 Possible rational roots: Find f(1), f(-1), f(2), and f(-2). Next find f(-2). f(-2) = (-2)3 - 2(-2)2 - 1(-2) + 2 f(-2) = -8 - 8 + 2 + 2 f(-2) = {1, -1, 2, -2} ? ✅ -2 is not a root of the function -2 is a root of the function Select the true statement. ✅ Using the function f(x) = x3 - 2x2 - 1x + 2 Possible rational roots: Find f(1), f(-1), f(2), and f(-2). Finally, find f(2). f(2) = (2)3 - 2(2)2 - 1(2) + 2 f(2) = 8 - 8 - 2 + 2 f(-2) = {1, -1, 2, -2} ? ✅ -2 is not a root of the function -2 is a root of the function Select the true statement. ✅ ❌ Using the function f(x) = x3 - 2x2 - 1x + 2 Actual rational roots: {1, -1, 2, -2} ? Hit OK and you're done! ❌ |