Maths

32 MATHEMATICS
+ ✁ + ✂ = –b
a ,
✁ + ✁✂ + ✂ = c a ,
Let us consider an example.
✁ ✂ = – d
a .
1
Example 5* : Verify that 3, ✄ –1, are the zeroes of the cubic polynomial
3
p(x) = 3x 3 – 5x 2 – 11x – 3, and then verify the relationship between the zeroes and the
coefficients.
Solution : Comparing the given polynomial with ax 3 + bx 2 + cx + d, we get
a = 3, b = – 5, c = –11, d = – 3. Further
p(3) = 3 × 3 3 – (5 × 3 2 ) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,
p(–1) = 3 × (–1) 3 – 5 × (–1) 2 – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,
3 2
1 1 1 1
3 5 11 3,
3 3 3 3
p ✆ ☎ ✆ ☎ ✆ ☎ ✆
✝ ✞ ✟ ✝ ✝ ✟ ✝ ✝ ✟ ✝ ✝
☎ ✡ ✠ ✡ ✠ ✡ ✠ ✡
✠
☛ ☞ ☛ ☞ ☛ ☞ ☛ ☞
=
1 5 11 2 2
✄ – ✌ 3 ✄ – ✍ 0 ✌ ✍
9 9 3 3 3
1
Therefore, 3, –1 and are the ✄ zeroes of 3x
3
3 – 5x 2 – 11x – 3.
1
So, we ✂ take = ✄ ✎
3, = –1 and ✁ =
3
Now,
1 ✏ 1 ✏ 5 ✏ ( ✑
5)
✒
b
3 ( ✓ ✔ ✕ ✔ ✖ ✗ ✔ ✏ ✔ ✏ ✗ ✏ ✗ ✗ ✗
1) 2
,
3 3 3 3 a
✘
✚
✙
✛
1 1 ✏
✑ ✒ ✑ 1 ✒
11
3 ( 1) ( 1) 3 3 1
3 3 3 3
✔ ✕ ✖ ✔ ✖ ✓ ✗ ✜ ✏ ✔ ✏ ✜ ✏ ✔ ✏ ✜ ✗ ✏ ✔ ✏ ✗ ✗
✘ ✙ ✘ ✙
✓✕
✛ ✚ ✛
✚
c
a
,
1 ( 3) ✢ ✢
✤ ✣
3 ( 1) 1
3 3
✦ ✧ ★ ✩ ✢ ✩ ✢ ★ ★ ★
✪ ✫
✥
d ✢
a
.
* Not from the examination point of view.
✬
✭