Short Answer Type
Q1: If a and b are roots of the equation x^{2} + 7 x + 7 . Find the value of a^{−1} + b^{−1} − 2αb ?
Ans: for f ( x ) = x^{2} + 7 x + 7
we get
α + b = − 7
αb = 7
Now a^{−1} + b^{−1} − 2αb
Q2: If the zeroes of the quadratic polynomial x^{2} + (α + 1 ) x + b are 2 and 3, then find the value of a and b.
Ans: Let f (x) = x^{2} + (a + 1) x + b
Then 2 − 3 = − ( a + 1 ) or α = 0
− 6 = b
So a = 0 and b = 6
Q.3. If a and b are zeroes of the polynomial f (x) = 2x^{2} − 7x + 3, find the value of α^{2} + b^{2}.
Ans: f (x) = 2x^{2} − 7 x + 3
= 2x^{2} − x − 6 x + 3 = x(2x − 1) − 3(2 x − 1) = (x − 3) (2 x − 1)
So zeroes are 3 and 1/2
Now
α^{2} + b^{2}
Q.4. Find the zeroes of the quadratic polynomial x^{2} + x − 12 and verify the relationship between the zeroes and the coefficients.
Ans: x^{2} + x − 12
= x^{2} + 4x − 3x − 12
= x (x + 4) − 3 (x + 4) = (x − 3) (x + 4)
So zeroes are 3 and 4
as we know
sum of roots = b/a = (1)/1 = 1 ie 3+(4) = 1
product of roots = c/a = 12/1 = 12 ie 3x4 = 12
Q5: If p and q are zeroes of f (x) = x^{2} − 5x + k, such that p − q = 1 , find the value of k.
Ans: for f ( x ) = x^{2} − 5 x + k
we get p + q = 5
pq = k
Now p − q = 1
(p − q)^{2} = 1
(p + q)^{2} − 4pq = 1
25 − 4k = 1
k = 6
Q6: Given that two of the zeroes of the cubic polynomial αx^{3} + bx^{2} + cx + d are 0, then find the third zero.
Ans: Two zeroes = 0, 0
Let the third zero be k.
The, using relation between zeroes and coefficient of polynomial, we have:
k + 0 + 0 = − b a
Third zero = k = b/a
Q.7. If one of the zeroes of the cubic polynomial x^{3} + αx^{2} + bx + c is 1, then find the product of the other two zeroes.
Ans:
Q8: If ab, a a+b , are zeroes of x^{3} − 6x^{2} + 8x , then find the value of b
Ans: Let f ( x ) = x^{3} − 6x^{2} + 8 x
Method 1
= x (x^{2} − 6 x + 8) = x(x − 2) (x − 4)
So 0,2,4 are zeroes of the polynomial. or a=2 and b=2 or 2 Method 2
Q9: Quadratic polynomial 4x^{2} + 12x + 9 has zeroes as p and q . Now form a quadratic polynomial whose zeroes are p − 1 and q − 1
Ans: 4x^{2} + 12 x + 9
= 4x^{2} + 6x + 6x + 9
= 2x(2x + 3) + 3(2x + 3) = (2 x + 3)^{2}
So p = 3/2 and q = 3/2
So, p − 1 = − 5/2 and q − 1 = − 5/2
So quadratic polynomial will be
(x + 5/2)^{2}
or
4x^{2} + 20 x + 25
Long Answer Type
Q10: p and q are zeroes of the quadratic polynomial x^{2} − (k + 6 ) x + 2(2 k − 1) . Find the value of k if 2(p + q) = p q
Ans: for f (x) = x^{2} − (k + 6) x + 2 (2 k − 1)
We get, p + q = k + 6
p q = 2(2 k − 1)
Now 2 (p + q ) = pq
Therefore,
2 (k + 6) = 2(2k − 1)
or k + 6 = 2 k − 1
or k = 7
Q11: Given that the zeroes of the cubic polynomial x^{3} − 6 x^{2} + 3 x + 10 are of the form a, a + b, a + 2b for some real numbers a and b, find the values of a and b as well as the zeroes of the given polynomial.
Ans: k_{1} + k_{2} + k_{3} =
a + a + b + a + 2b = 6
a + b = 2
a = 2 − b
Now
b=3 or b=3
So a= 5 or 1
The zeroes with a = 5, b= 3 can be expressed as 5, 2, 1
The zeroes with a = 1, b = 3 can be expressed as 1, 2, 5
Q12: If one zero of the polynomial 2x^{2}−5x−(2k + 1) is twice the other, find both the zeroes of the polynomial and the value of k.
Ans: Let a be one zero ,then another will be 2a
Now
α + 2α = 5/2 or a= 5/6
Also
k= 17/9
Q.13: Using division show that 3y^{2} + 5 is a factor of 6y^{5} + 15y^{4} + 16y^{3} + 4y^{2} + 10y − 35 .
Ans:
Q14: If (x  2) and [x  1/2 ] are the factors of the polynomials qx^{2} + 5x + r prove that q = r.
Ans: 4q + 10 + r = 0 (1)
q/4 +5/2 + r = 0 or q + 10 + 4r = 0 (2)
Subtracting 1 from 2
3q3r = 0
q = r
Q15: Find k so that the polynomial x^{2} + 2x + k is a factor of polynomial 2x^{4} + x^{3}  14x^{2} + 5x + 6. Also, find all the zeroes of the two polynomials.
Ans: For x^{2} + 2x + k is a factor of polynomial 2x^{4} + x^{3}  14x^{2} + 5x + 6, it should be able to divide the polynomial without any remainder
Comparing the coefficient of x we get.
21 + 7k = 0
k = 3
So x^{2} + 2x + k becomes x^{2} + 2x 3 = (x1)(x+3)
Now
2x^{4} + x^{3}  14x^{2} + 5x + 6= (x^{2} + 2x 3)(2x^{2}3x8+2k)
=(x^{2} + 2x  3)(2x^{2}3x  2)
=(x  1)(x + 3)(x  2)(2x + 1)
or x = 1, 3, 2,= 1/2
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1. What are polynomials and how are they classified? 
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