1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; 1/2 , 1, -2
(ii) x3 – 4x2 + 5x – 2; 2,1,1
Solution:
(i) 2x3 + x2 – 5x + 2; 1/2 , 1, -2
p(x) = 2x3 + x2 – 5x + 2
Zeros for this polynomial are 1/2 , 1, -2
p(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2
= (1/4) + (1/4) – (5/2) + 2
= 0
p(1) = 2(1)3 + (1)2 – 5(1) + 2 = 0
p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2
= -16 + 4 + 10 + 2
= 0
Therefore, ½ , 1, and −2 are the zeroes of the given polynomial. Comparing the given polynomial with ax3 + bx2 + cx + d, we obtain a = 2, b = 1, c = −5, d = 2
Let us take α = 1/2 ,β = 1, γ = -2
α + β + γ = 1/2 + 1 + (-2) = –1/2 = –b/a
αβ + βγ + γα = 1/2 x 1 + 1(-2) + 1/2 (-2) = -5/2 = c/a
αβγ = 1/2 x 1 x (-2) = –1/1 = – 2/2 = –d/a
Therefore, the relationship between the zeroes and the coefficients is verified.
(ii) x3 – 4x2 + 5x – 2; 2,1,1
p(x) = x3 – 4x2 + 5x – 2
Zeros for this polynomial are 2 , 1, 1
p(2) = 23 – 4(2)2 + 5(2) – 2
= 8 – 16 + (10) – 2
= 0
p(1) = (1)3 – 4(1)2 + 5(1) – 2 = 0
Therefore, 2 , 1, and 1 are the zeroes of the given polynomial. Comparing the given polynomial with ax3 + bx2 + cx + d, we obtain a = 1, b = -4, c = 5, d = -2
Verify:
Sum of zeros = 2+1+1 = 4 = – (-4)/1 = –b/a
Multiplication of zeroes taking two at a time = (2)(1)+(1)(1)+(2)(1) = 2+1+2= 5 = 5/1 = c/a
Multiplication of zeroses = 2 x 1 x 1 = 2 = – (-2)/1 = –d/a
Hence, the relationship between the zeroes and the coefficients is verified.
2. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.
Solution:
Let the polynomial be ax3 + bx2 + cx + d and the zeroes be α, β and γ.
It is given that
α + β + γ = 2/1 = –b/a
αβ + βγ + γα = -7/2 = c/a
αβγ = – 14/1 = –d/a
If a = 1, then b = −2, c = −7, d = 14 Hence, the polynomial is x3 – 2x2 – 7x + 14
3. If the zeroes of polynomial, x3 -3x2+x+1 are a-b,a,a+b. Find a and b.
Solution:
p(x) = x3 -3x2+x+1
Zeroes are a − b, a + a + b Comparing the given polynomial with px3+qx2+rx+1, we obtain p = 1, q = −3, r = 1, t = 1
Sum of zeros = a – b + a + a + b
-q/ p = 3a
–(-3)/1 = 3a
3 = 3a
a = 1
The zeroes are 1-b, 1, 1+b
Multiplication of zeroes = 1(1-b)(1+b)
-t/p = 1 – b2
-1/1= 1 – b2
1 – b2 = -1
1 + 1 = b2
b = ±√2
4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ±√3 , find other zeroes.
Solution:
Given 2 + √3 and 2 – √3 are zeroes of the given polynomial. So, (2 + √3)(2 – √3) is a factor of polynomial.
Therefore,[x – (2 + √3)][x – (2 – √3)] = x2 + 4 − 4x − 3 = x2 − 4x + 1 is a factor of the given polynomial
For finding the remaining zeroes of the given polynomial, we will find
Clearly, x4 -6x3 -26x2+138x -35 = (x2 – 4x + 1)(x2 – 2x – 35)
We know, (x2 – 2x – 35) = (x – 7)(x + 5)
Therefore, the value of the polynomial is also zero when (x -7 = 0) or (x + 5 = 0)
Or x = 7 or −5
therefore, 7 and −5 are also zeroes of this polynomial.
5. If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:
By division algorithm we know,
Dividend = Divisor × Quotient + Remainder
Dividend − Remainder = Divisor × Quotient
x4 – 6x3 + 16x2 – 25x + 10 – x – a = x4 – 6x3 + 16x2 – 26x + 10 – a will be divisible by x2 – 2x + k
It ccan be observed that (-10+2k)x +(10-a-8k+k2) will be zero.
Thus, (-10 + 2k) = 0 and (10 – a – 8k +k2) = 0
For (-10+2k) = 0 , 2k = 10 and thus k = 5
For (10 – a – 8k + k2) = 0
10 − a − 8 × 5 + 25 = 0
10 − a − 40 + 25 = 0
− 5 − a = 0
Therefore, a = −5
Hence, k = 5 and a = −5
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