Rational Root Theorem and Fundamental Theorem of Algebra

• Yesterday we were factoring polynomials using synthetic division but we were given a place to start. How would we get started if we were not given a factor?

You have learned several important properties about real roots of polynomial equations.

Not all polynomials are factorable, but the Rational Root Theorem can help you find all possible rational roots of a polynomial equation.

Example 2: Finding All Roots of a Polynomial Solve x 4 – 3 x 3 + 5 x 2 – 27 x – 36 = 0 by finding all roots. The polynomial is of degree 4, so there are exactly four roots for the equation. Step 1 Use the rational Root Theorem to identify rational roots. p = – 36, and q = 1. ± 1, ± 2, ± 3, ± 4, ± 6, ± 9, ± 12, ± 18, ± 36

Example 2 Continued Step 2 Graph y = x 4 – 3 x 3 + 5 x 2 – 27 x – 36 to find the real roots. Find the real roots at or near – 1 and 4.

Example 2 Continued Step 3 Test the possible real roots. – 1 – 3 – 1 5 4 – 9 36 1 – 4 9 – 36 0 1 – 27 – 36 Test – 1. The remainder is 0, so (x + 1) is a factor.

Example 2 Continued The polynomial factors into (x + 1)(x 3 – 4 x 2 + 9 x – 36) = 0. 4 1 – 4 4 9 0 – 36 36 1 0 9 0 Test 4 in the cubic polynomial. The remainder is 0, so (x – 4) is a factor.

Example 2 Continued The polynomial factors into (x + 1)(x – 4)(x 2 + 9) = 0. Step 4 Solve x 2 + 9 = 0 to find the remaining roots. x 2 + 9 = 0 x 2 = – 9 x = ± 3 i The fully factored form of the equation is (x + 1)(x – 4)(x + 3 i)(x – 3 i) = 0. The solutions are 4, – 1, 3 i, – 3 i.

Check It Out! Example 2 Solve x 4 + 4 x 3 – x 2 +16 x – 20 = 0 by finding all roots. The polynomial is of degree 4, so there are exactly four roots for the equation. Step 1 Use the rational Root Theorem to identify rational roots. ± 1, ± 2, ± 4, ± 5, ± 10, ± 20 p = – 20, and q = 1.

Check It Out! Example 2 Continued Step 2 Graph y = x 4 + 4 x 3 – x 2 + 16 x – 20 to find the real roots. Find the real roots at or near – 5 and 1.

Check It Out! Example 2 Continued Step 3 Test the possible real roots. – 5 1 4 – 1 16 – 20 – 5 5 – 20 20 1 – 1 4 – 4 0 Test – 5. The remainder is 0, so (x + 5) is a factor.

Check It Out! Example 2 Continued The polynomial factors into (x + 5)(x 3 – x 2 + 4 x – 4) = 0. 1 1 – 1 1 4 – 4 0 4 1 4 0 0 Test 1 in the cubic polynomial. The remainder is 0, so (x – 1) is a factor.

Check It Out! Example 2 Continued The polynomial factors into (x + 5)(x – 1)(x 2 + 4) = 0. Step 4 Solve x 2 + 4 = 0 to find the remaining roots. x 2 + 4 = 0 x 2 = – 2 x = ± 2 i The fully factored form of the equation is (x + 5) (x – 1)(x + 2 i)(x – 2 i) = 0. The solutions are – 5, 1, – 2 i, +2 i).

YOU Try Find the roots for the function… X+1 is NOT factor X-2 IS A factor X-1 is NOT a factor 18

You try 4. Solve by finding all roots. x 4 – 5 x 3 + 7 x 2 – 5 x + 6 = 0 •