Question

Given the equation x4 + 4x2 – 5x² – 16x +4 = 0, complete the following. a. List all possible rational roots. b. Use synthetic division to test several possible rational roots in order to identify one actual root. c. Use the root from part (b) to solve the equation. a. List all rational roots that are possible according to the Rational Zero Theorem. (Use commas to separate answers as needed.) b. Use synthetic division to test several possible rational roots in order to identify one actual root. One rational root of the given equation is (Simplify your answer.) c. Use the root from part (b) to solve the equation. The solution set is O (Simplify your answer. Type an exact answer, using radicals as needed. USe integers or fractions for any numbers in the expression. Use commas to separate answers as needed.)

Answer 1

Solutim: Given, polynomial is - *4 +473-574-167+4 = 0 Factory of constant term are : † 1, 12 Factory of leading coefficient are : Therefore, by rationale zero theorem, all possible satiệnal zeros 4 +1 ame +4. + 1 11 32 £1 £1 cie', I 1, 12, 4 > All possible rational zeros are ; #1, #2, 14 (b) Using Synthetic division method - 1 1 4 -16 4 1 olur vi -16 1 -16 -12 1 is not an actual root of here it gives remainder -12. 1 4 -5 boles in -3 Ô 4 a 12 1 3 1 is not an actual root as it geves semaines 12. 1 4 4 -5 12 7 -16 14 1 --- 6 2. is an actual root. 1 -5 G -16 10 4 2 -9 :: → -2 is also an actual root

1 -16 4 4 16 20 -5 80 75 300 4 1136 1140 4 284 4 mot" an actual root as it geves remainder 1140. -4 1 4 -5 -16 4. -16 20 1 #10 oli -5 4 -12 -4 is not an actual root as it gives remainder -12. Therefore, I rational roats which are actual reats are : 12 Ame. (c) since, ta and ~a are roots of given polynomial, therefore (x+2)(7-2) must dinilete the given polynomial. x?_ 4) ** + 4x3_522_16x +473?+42–1 28 photos 4742 + 4x²-x2 473 – 16:4 + -X? +4 -X+4. → **+ 482 - 5** – 26% +4 = (x+2)(2-2). (?+ 4x-1) Now, 20+43-1 = 0 i -4+]42-4X 1XE17 → fiurering Fridhare changes 2X.1 formula 2= -27 55 Therefore, the solution sete is : 12,-2+5 → The Solution set is : +2, -2, -4+15) -2-55 Ang.

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