Sunday, December 11, 2016

Chapter 20.4 - Polynomials - Solved examples

In the previous section we completed the discussion on polynomials. In this section we will see a few solved examples based on our discussion in this chapter as a whole.

Solved example 20.16
Given that:
• p(x) = ax+ bx2 + cx + d, 
• p(-2) = 0. 
Prove that 8a + 2c = 4b + d
Solution:
We have p(x) = ax+ bx2 + cx + d
So p(-2) = a×(-2)3 + b×(-2)2 + c×(-2) + d = 0 
⇒ -8a + 4b -2c +d = 0
⇒ 8a + 2c = 4b + d

Solved example 20.17
Given that:
• p(x) = x3 - 4x + 3 
• q(x) = 2x3 - 3x2 + 11
Which polynomial should be subtracted from [p(x) + q(x)] to get (x3 - x2 - 5)
Solution:
First we will calculate [p(x) + q(x)]: 
[x3 - 4x + 3] + [2x3 - 3x2 + 11] = 3x3 - 3x2 - 4x + 14
Let s(x) be the required polynomial. That means:
When we subtract s(x) from [p(x) + q(x)], we will get (x3 x2 - 5). So we can write:
[p(x) + q(x)] - s(x) = (x3 x2 - 5) 
 (3x3 - 3x2 - 4x + 14) -s(x) = (x3 x2 - 5
 s(x) = (3x3 - 3x2 - 4x + 14) - (x3 x2 - 5)
 s(x) = (2x3 - 2x2 - 4x + 19)

Solved example 20.18
Given that:
• p(x) = 2x+ 9x2 + kx + 3 
• p(-2) = p(-3) 
What is the value of k?
Solution:
1. We have p(x) = 2x+ 9x2 + kx + 3 
2. p(-2) = 2×(-2)3 + 9×(-2)2 + k×(-2) + 3 = -16 + 36 -2k +3 = 23 -2k
3. p(-3) = 2×(-3)3 + 9×(-3)2 + k×(-3) + 3 = -54 + 81 -3k +3 = 30 -3k
4. Given that these two are equal So we can write:
23-2k = 30-3k 
 3k-2k = 30 -23 
⇒ k = 7

Solved example 20.19
Given that:
• p(x) = ax+ bx2 + cx + d 
• p(1) = p(-1) 
Prove that a + c = 0
Solution:
1. We have p(x) = ax+ bx2 + cx + d
2. p(1) = a×(1)3 + b×(1)2 + c×(1) + d = a +b + c + d
3. p(-1) = a×(-1)3 + b×(-1)2 + c×(-1) + d = -a + b -c +d
4. Given that these two are equal So we can write:
+b + c + d =  -a + b - c +d
 2a + 2c = 0
⇒ a + c = 0

Solved example 20.20
Given that:
• p(x) = 2x+ ax2 - 7x + b 
• p(1) = 3 
• p(2) = 19
What is the value of a and b?
Solution:
1. We have p(x) = 2x+ ax2 - 7x + b 
2. p(1) = 2×(1)3 + a×(1)2 - 7×(1) + b = 2 + a -7 + b = 3
 a + b -5 = 3  a + b = 8
3. p(2) = 2×(2)3 + a×(2)2 - 7×(2) + b = 16 + 4a - 14 + b = 19
⇒ 4a + b + 2 = 19 ⇒ 4a + b = 17 
4. From (2) we get: b = (8-a)
5. Substituting this value of b in (3) we get:
4a + (8-a) = 17 ⇒ 3a + 8 = 17 ⇒ 3a = 9 ⇒ a = 3
6. Substituting this value of a in (4) we get: b = 8 - 3 = 5

Solved example 20.21
If the degree of p(x) is 5 and the degree of q(x) is 3, what is the degree of [p(x) + q(x)]?
Solution:
1. Given that, the degree of p(x) is 5. It means that the 'leading term' in p(x) have an exponent 5. This term will be present in [p(x) + q(x)]
2. Given that, the degree of q(x) is 3. It means that the 'leading term' in q(x) have an exponent 3. This term will be present in [p(x) + q(x)]
3. In [p(x) + q(x)], there will not be any terms with an exponent greater than 5
4. So the leading term in [p(x) + q(x)] will have an exponent 5
5. Thus the degree of [p(x) + q(x)] is 5

Solved example 20.22
If the degree of [p(x) × (3x2 - 4x + 2)] is to be 5, What should be the degree of p(x)?
Solution:
1. The degree of p(x) is the exponent of it's leading term. Let this be 'n'
2. The degree of (3x2 - 4x + 2) is 2, which is the exponent of 3x2 
3. The degree of [p(x) × (3x2 - 4x + 2)] will be the exponent of the product:
[leading term of p(x)]  ×  [leading term of (3x2 - 4x + 2)]
4. The exponent of this product will be (n+2)
5. We want (n+2) = 5
6. So n should be equal to 3
7. That means, the degree of p(x) should be 3

Solved example 20.23
What should be added to (2x+ 3x2 + 8x - 4)  to get 0?
Solution:
1. Let p(x) be the required polynomial. That means:
2. When we add p(x) to (2x+ 3x2 + 8x - 4), we will get 0. So we can write:
(2x+ 3x2 + 8x - 4) + p(x) = 0
⇒ p(x) = 0 - (2x+ 3x2 + 8x - 4= (-2x- 3x2 - 8x + 4)

In the next section we will see Circle Measures.


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