DO THIS
If α,β, y are the zeroes of the given cubic polynomials, find the values of the expressions
given in the table.
| s.No | Cubic Polynomial | α+β+ y | αβ + βy+ yα | αβy |
| 1 | x3
+ 3x
2
– x – 2 | | | |
| 2 | 4x
3
+ 8x
2
– 6x – 2 | | | |
| 3 | x3
+ 4x
2
– 5x – 2 | | | |
| 4 | x3
+ 5x
2
+ 4 | | | |
Let us consider an example.
Example-7. Verify whether 3, –1 and –
1/
3
are the zeroes of the cubic polynomial
p(x) = 3x
3
– 5x2
– 11x – 3, and then verify the relationship between the zeroes and the coefficients.
Solution : p(x) = 3x
3
– 5x
2
– 11x – 3 is the given polynomial.
Then p(3) = 3 × 33
– (5 × 32
) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1)3
– 5 × (–1)2
– 11 × (–1) – 3 = – 3 – 5 + 11 – 3 = 0,
p(-1/3)=3x(-1/3)3-5x(-1/3) 2 -11x(-1/3)-3,
=-1/9-5/9+11/3-3=-2/3+2/3=0
Therefore, 3, –1, and
-1/
3
, are the zeroes of 3x
3– 5x
2 – 11x – 3
So, we take α = 3, β = –1 and y =
-1/
3
, 0
Comparing the given polynomial with ax3 bx2 + cx + d, we get
a= 3, b = – 5, c = –11, d = – 3.
Now,
α + β+ y = 3 + (–1) +
[-1/3
]
= 2
-1/
3
=
5/
3
=
-(-5)/
3
=
-b/
a
,
αβ + βy + yα = 3 × (–1) + (–1) × -[1/3]
+
[1/
3]
× 3 = – 3 +
1/
3
– 1 =
-11/
3
,
=
c/
a
,
αβy = 3 × (–1) x[-1/3] = 1 =-(-3)/3
=
-d/a.
page no:70
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