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Factorise by the division of polynomials: x^3-5x^2+2x+24÷x-4 |
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Answer» Answer: Let ( ) = x 3 − 5 x − 2 x + 24
By the rational root theorem, any rational zeros of f ( x ) must be expressible in the for p q for integers p , q with p a divisor of the constant term 24 and q a divisor of the coefficient 1 of the leading term. That means that the only possible rational zeros are the factors of 24 , namely: ± 1 , ± 2 , ± 3 , ± 4 , ± 6 , ± 12 , ± 24
Try each in turn: f ( 1 ) = 1 − 5 − 2 + 24 = 18
f ( − 1 ) = − 1 − 5 + 2 + 24 = 20
f ( 2 ) = 8 − 20 − 4 + 24 = 8
f ( − 2 ) = − 8 − 20 + 4 + 24 = 0
So x = − 2 is a zero and ( x + 2 ) is a factor. x 3 − 5 x 2 − 2 x + 24 = ( x + 2 ) ( x 2 − 7 x + 12 ) We can factor x 2 − 7 x + 12 by NOTING that 4 × 3 = 12 and 4 + 3 = 7 , so: x 2 − 7 x + 12 = ( x − 4 ) ( x − 3 )
Putting it all together: x 3 − 5 x 2 − 2 x + 24 = ( x + 2 ) ( x − 4 ) ( x − 3 ) |
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