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Question
2x4 − 7x3 − 13x2 + 63x − 45
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Solution
Let f(x) 2x4 − 7x3 − 13x2 + 63x − 45 be the given polynomial.
Now, putting x = 1we get
`f(1) = 2(1)^4 - 7(1)^3 - 13(1)^2 + 63(1) - 45`
` = 2-7 -13 + 63 - 45`
` = 65 - 65 = 0`
Therefore,(x -1)is a factor of polynomial f(x).
Now,
`f(x) = 2x^3 (x -1) - 5x^2(x -1) - 18x (x-1) + 45(x -1)`
` = (x -1) {2x^3 - 5x^2 - 18x + 45}`
` = (x -1)g(x) .... (1)`
Where `g(x) = 2x^2 - 5x^2 - 18x + 45`
Putting x = 3,we get
`g(3) = 2(3)^3 - 5(3)^2 - 18(3) + 45`
` = 54 - 45 - 54 + 45`
` = 0`
Therefore, (x -3)is a factor of g(x).
Now,
`g(x) = 2x^2 (x -3) + x(x -3) - 15(x -3)`
` = (x -3){2x^2 + x - 15}`
` = (x -3){2x ^2 + 6x - 5x - 15}`
` = (x -3){(2x - 5)(2x +3)}`
` = (x - 3) (x + 3)(2x -5) ..............(2)`
From equation (i) and (ii), we get
`f(x) = (x -1) (x -3)(x +3) (2x -5)`
Hence (x - 1),(x - 3),(x + 3) and (2x-5)are the factors of polynomial f(x).
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