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Question
It is given that –1 is one of the zeroes of the polynomial x3 + 2x2 – 11x – 12. Find all the zeroes of the given polynomial.
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Solution
Let f(x) = x3 + 2x2 – 11x – 12
Since –1 is a zero of f(x), (x + 1) is a factor of f(x).
On dividing f(x) by (x + 1), we get
`x + 1")"overline(x^3 + 2x^2 - 11x - 12)"("x^2 + x + 12`
x3 + x2
– –
x2 – 11x – 12
x2 + x
– –
–12x – 12
–12x – 12
+ +
X
f(x) = x3 + 2x2 – 11x – 12
= (x + 1) (x2 + x – 12)
= (x + 1) {x2 + 4x – 3x – 12}
= (x + 1) {x(x + 4) – 3(x + 4)}
= (x + 1) (x – 3) (x + 4)
∴ f(x) = 0
⇒ (x + 1) (x – 3) (x + 4) = 0
⇒ (x + 1) = 0 or (x – 3) = 0 or (x + 4) = 0
⇒ x = – 1 or x = 3 or x = – 4
Thus, all the zeroes are – 1, 3 and – 4.
