Lesson Rational Root Theorem Part 2
 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 denominatorThe 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 - 6Factors of -6?Factors of 2? Write an expression that represents all possible rationalroots of the polynomial: f(x) = 2x3 - 5x2 + x - 6Factors of -6Factors of 2±±1, 2, 3, 6?1, 2? There are three ways.How do we know which numbers on the listIf 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 division3. Graph the related functionare actually answers/roots/solutions? Using the function f(x) = x3 - 2x2 - 1x + 2±±1,  2?1?let's find the list of possible rational rootsand see which ones actually workPossible rational roots: {1, -1, 2, -2}? Using the function f(x) = x3 - 2x2 - 1x + 2Possible 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) + 2f(1) = 1 - 2 - 1 + 2f(1) = {1, -1, 2, -2}1 is not a root of the function1 is a root of the functionSelect the true statement. Using the function f(x) = x3 - 2x2 - 1x + 2Possible 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) + 2f(-1) = -1 - 2 + 1 + 2f(-1) = {1, -1, 2, -2}?✅-1 is not a root of the function-1 is a root of the functionSelect the true statement. Using the function f(x) = x3 - 2x2 - 1x + 2Possible 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) + 2f(-2) = -8 - 8 + 2 + 2f(-2) = {1, -1, 2, -2}?✅-2 is not a root of the function-2 is a root of the functionSelect the true statement.✅ Using the function f(x) = x3 - 2x2 - 1x + 2Possible 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) + 2f(2) = 8 - 8 - 2 + 2f(-2) = {1, -1, 2, -2}?✅-2 is not a root of the function-2 is a root of the functionSelect the true statement.✅❌ Using the function f(x) = x3 - 2x2 - 1x + 2Actual rational roots: {1, -1, 2, -2}?Hit OK and you're done!❌
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