TÍTULO IX Infracciones, sanciones y recursos
2.2 LA EXPORTACION EN MEXICO
Introduction:
Degree of a Polynomial:
z If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial p(x).
Linear polynomial:
z A polynomial of degree 1 is called a linear polynomial.
z For example, 2x – 3, 3 x + 5, y + 2, x – 11
2 , 3z + 4, 3
2u + 1 etc. are linear polynomials.
z General form of linear polynomial is ax + b, where a, b are real numbers and a ≠ 0.
Quadratic polynomial:
z A polynomial of degree 2 is called a quadratic polynomial.
z The name ‘quadratic’ has been derived from the word ‘quadrate’, which means ‘square’.
z 2x2 + 3x – 5
2, y2 – 2, 2 – x2 + 3 x, 3
u − 2u2 + 5, 5 v2 – 3
2v, 4z2 + 7
1 are some examples of quadratic polynomials (whose coefficients are real numbers).
z Generally, any quadratic polynomial in x is of the form ax2 + bx + c, where a, b, c are real numbers and a ≠ 0.
Cubic polynomial:
z A polynomial of degree 3 is called a cubic polynomial.
z Some examples of a cubic polynomial are 2 – x3, x3, 2x3, 3 – x2 + x3, 3x3 – 2x2 + x – 1.
z In fact, the most general form of a cubic polynomial is ax3 + bx2 + cx + d, where, a, b, c, d are real numbers and a ≠ 0.
Value of a polynomial:
z If p(x) is a polynomial in x, and if k is any real number, then the value obtained by replacing x by k in p(x), is called the value of p(x) at x = k, and is denoted by p(k).
z Consider the polynomial P(x) = x2 – 3x – 4 value of the polynomial at x = 3 is denoted by P(3) and P(3) = 32 – 3(3) – 4 = –4.
z Value of the above polynomial at x = –1 is given by P(–1) = (–1)2 – 3(–1) – 4 = 0.
→ Introduction
→ Geometrical meaning of Zeroes of a Polynomial
→ Relationship between zeroes and coefficients of a Polynomial
→ Division Algorithm for Polynomials
Chapter 02: Polynomials 17
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 17
Zero of a polynomial:
z A real number k is said to be a zero of a polynomial p(x), if p(k) = 0.
z If x – a is a factor of the polynomial p(x), then p(a) = 0
z If x + a is a factor of the polynomial p(x), then p(–a) = 0
z Consider the polynomial P(x) = x2 – 5x + 6 P(2) = 22 – 5(2) + 6 = 0 and
P(3) = 32 – 5(3) + 6 = 0
z As P(2) = 0 and P(3) = 0, 2 and 3 are called the zeros of the polynomial x2 – 5x + 6.
Geometrical Meaning of the Zeroes of a Polynomial:
¾ Now, let us look for the geometrical meaning of a zero of a quadratic polynomial.
¾ Consider the quadratic polynomial x2 – 3x – 4.
¾ Let us see what the graph of y = x2 – 3x – 4 looks like.
¾ Let us list a few values of y = x2 – 3x – 4 corresponding to a few values for x as given in Table.
x –2 –1 0 1 2 3 4 5
y = x2 – 3x – 4 6 0 –4 –6 –6 –4 0 6
Note from the figure that the curve intersect x–axis at the points –1 and 4. Thus the zeros of the Polynomial x2 – 3x – 4 are –1 and 4.
¾ From our observation earlier about the shape of the graph of y = ax2 + bx + c, the following three cases can happen:
¾ Case (i) : Here, the graph cuts x-axis at two distinct points A and A′.
¾ The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case (see given below).
Y
Y′
X′ 0 A A′ X
Y
Y′
X′ A′ 0 A X
(i) (ii) –3 –2 –1 0 1 2 3 4 5
1 2 3 4 5 6
–1 –2 –3 –4 –5 –6
(1,–6) (2,–6) (3,–4) (0,–4)
(4,0) (–1,0)
(–2,6) (5, 6)
18 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume z Case (ii) : Here, the graph cuts the x-axis at exactly one point, i.e. at two coincident points.
So, the two points A and A′ of Case (i) coincide here to become one point A (see Fig. given below).
z The x-coordinate of A is the only zero for the quadratic polynomial ax2 + bx + c in this case.
z Case (iii) : Here, the graph is either completely above the axis or completely below the x-axis. So, it does not cut the x-axis at any point (see Fig. given below).
z So, the quadratic polynomial ax2 + bx + c have no zero in this case.
Number of Zeros of a Polynomial:
z Given a polynomial P(x) of degree n, the graph of y = P(x) intersects the x–axis at most n points. Therefore a polynomial P(x) of a degree n has at most n zeros.
z That is, a quadratic polynomial can have at most two zeros and so on.
SOLVED EXAMPLE 2.1:
1) Look at the graphs in Fig. given below. Each is the graph of y = p(x), where p(x) is a polynomial. For each of the graphs, find the number of zeroes of p(x).
Y
Y′
X′ 0 X
A
Y
Y′
X′ 0 A X
(i) (ii)
Y
Y′
X′ 0 X
Y
Y′
X′ 0 X
(i) (ii)
Y
Y′
X′ 0 X
Y
Y′
X′ 0 X
(i) (ii) Y
Y′
X′ 0 X
(ii)
Chapter 02: Polynomials 19
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 19 Sol: i) The number of zeroes is 1 as the graph intersects the x-axis at one point only.
ii) The number of zeroes is 2 as the graph intersects the x-axis at two points.
iii) The number of zeroes is 3.
iv) The number of zeroes is 1.
v) The number of zeroes is 1.
vi) The number of zeroes is 4.
UNSOLVED EXERCISE 2.1:
1) The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Relationship between Zeroes and Coefficients of a Polynomial:
¾ If α and β are the zeroes of the quadratic polynomial p(x) = ax2 + bx + c, a ≠ 0, then you know that x – α and x – β are the factors of p(x).
¾ Therefore, ax2 + bx + c = k(x – α) (x – β), where k is a constant = k[x2 – (α + β)x + αβ] = kx2 – k(α + β)x + kαβ = k[x2 – (sum of zero)x + (Product of zero)]
Y
Y′
X′ 0 X
Y
Y′
X′ 0 X
(i) (ii) Y
Y′
X′ 0 X
(ii) Y
Y′
X′ 0 X
Y
Y′
X′ 0 X
(iv) (v) Y
Y′
X′ 0 X
(vi) Y
Y′
X′ 0 X
Y
Y′
X′ 0 X
(iv) (v)
Y
Y′
X′ 0 X
(vi)
20 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume
¾ Comparing the coefficients of x2, x and constant terms on both the sides, we get a = k, b = – k(α + β) and c = kαβ.
SOLVED EXAMPLES 2.2:
1) Find the zeroes of the quadratic polynomial x2 + 7x + 10, and verify the relationship between the zeroes and the coefficients.
Sol: We have, x2 + 7x + 10 = (x + 2)(x + 5)
2) Find the zeroes of the polynomial x2 – 3 and verify the relationship between the zeroes and the coefficients.
Sol: Recall the identity a2 – b2 = (a – b)(a + b)
3) Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2, respectively.
Chapter 02: Polynomials 21
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 21 4) If α and β are zero of the Quadratic Polynomial f(x) = ax2 + bx + c, then evaluate (i) α2 + β2
UNSOLVED EXERCISE 2.2:
CW Exercise
1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) x2 – 2x – 8 ii) 4s2 – 4s + 1 iii) 3x2 – x – 4 iv) 7z2 – 343 v) 5p2 + 20p
2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. the zeros and its coefficients.
4) Find the zeros of the quadratic polynomial f(x) = abx2 + (b2 – ac) x – bc, and verify the relationship between the zeros and its coefficients.
5) If α and β are the zeros of the quadratic polynomial f(x) = x2 – px + q, then find the values of
1) Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
i) 6x2 – 3 – 7x [CBSE 08] ii) 4u2 + 8u iii) t2 – 15
22 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume 2) Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes
respectively.
i)
4
1, –1 ii) 0, 5 iii) 1, 1
3) If α and β are the zeros of the quadratic polynomial f(x) = x2 – x – 4, find the value of −αβ +β α
1
1 .
4) If α & β are the zeros of the quadratic polynomial P(x) = 4x2 –5x – 1, find the value of α2β + αβ2. 5) If α & β are the zeros of the polynomial f(x) = x2 – 5x + k such that α – β = 1, find the value of k.
6) If α, β are the zeros of the polynomial f(x) = 2x2 + 5x +k satisfying the relation α2 + β2 + αβ = 4 21, then find the value of k for this to be possible.
7) If sum of the square of the zeros of the quadratic polynomial f(x) = x2 – 8x + k is 40, find the value of k.
8) If α and β are the zeros of the quadratic polynomial f(x) = 2x2 – 5x + 7, find a polynomial whose zeros are 2α + 3β and 3α + 2β.
Division Algorithm for Polynomials:
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x).
This result is known as the Division Algorithm for polynomials.
i.e. Dividend = Divisor × Quotient + Reminder
SOLVED EXAMPLES 2.3:
1) Divide 3x3 + x2 + 2x + 5 by 1 + 2x + x2.
Sol: We first arrange the terms of the dividend and the divisor in the decreasing order of their degrees. Recall that arranging the terms in this order is called writing the polynomials in standard form. In this example, the dividend is already in standard form, and the divisor, in standard form, is x2 + 2x + 1.
Step 1: To obtain the first term of the quotient, divide the highest degree term of the dividend (i.e. 3x3) by the highest degree term of the divisor (i.e. x2). This is 3x. Then carry out the division process. What remains is – 5x2 – x + 5.
Step 2: Now, to obtain the second term of the quotient, divide the highest degree term of the new dividend (i.e. –5x2) by the highest degree term of the divisor (i.e. x2). This gives –5. Again carry out the division process with –5x2 – x + 5.
Step 3: What remains is 9x + 10. Now, the degree of 9x + 10 is less than the degree of the divisor x2 + 2x + 1. So, we cannot continue the division any further.
So, the quotient is 3x – 5 and the remainder is 9x + 10. Also,
(x2 + 2x + 1) × (3x – 5) + (9x + 10) = 3x3 + 6x2 + 3x – 5x2 – 10x – 5 + 9x + 10 = 3x3 + x2 + 2x + 5
Here again, we see that, Dividend = Divisor × Quotient + Remainder 2) Divide 3x2 – x3 – 3x + 5 by x – 1 – x2, and verify the division algorithm.
Sol: Note that the given polynomials are not in standard form. To carry out division, we first write both the dividend and divisor in decreasing orders of their degrees.
So, dividend = –x3 + 3x2 – 3x + 5 and divisor = –x2 + x – 1.
Division process is shown on the right side.
We stop here since degree (3) = 0 < 2 = degree (–x2 + x – 1).
So, quotient = x – 2, remainder = 3.
Chapter 02: Polynomials 23
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 23 Now, Divisor × Quotient + Remainder = (–x2 + x – 1) (x – 2) + 3
= –x3 + x2 – x + 2x2 – 2x + 2 + 3 = –x3 + 3x2 – 3x + 5 = Dividend In this way, the division algorithm is verified.
3) Find all the zeroes of 2x4 – 3x3 – 3x2 + 6x – 2, if you know that two of its zeroes are 2
and – 2. [CBSE–08]
Sol: Since two zeroes are 2 and − 2, (x – 2)(x + 2) = x2 – 2 is a factor of the given polynomial. Now, we divide the given polynomial by x2 – 2.
First term of quotient is 2 24 x
x = 2x2
Second term of quotient is 323 x
− x = –3x
Third term of quotient is 22 x x = 1
So, 2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2)(2x2 – 3x + 1).
Now, by splitting –3x, we factorise 2x2 – 3x + 1 as (2x – 1)(x – 1).
So, its zeroes are given by x = 2
1 and x = 1.
Therefore, the zeroes of the given polynomial are 2, – 2, 2
1 and 1.
UNSOLVED EXERCISE 2.3:
CW Exercise:
1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x 2) Check whether the first polynomial is a factor of the second polynomial by dividing the second
polynomial by the first polynomial:
i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12 ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
3) The graph of a polynomial f(x) = 3x4 + 6x3 – 2x2 – 10x – 5, intersects x axis at four different points P, Q, R and S. If the co-ordinates of points P and Q are ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ,0 3
5 and ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛− ,0 3 5
respectively then find the co-ordinates of R and S.
4) Polynomial f(x) = x3 – 2x – 4 on dividing by another polynomial g(x) gives equal quotient and remainder. Find out the value of g(x) if the remainder of the division is x-2.
5) Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and i) deg p(x) = deg q(x) ii) deg q(x) = deg r(x) iii) deg r(x) = 0
6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below.
i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x + 2 ii) f(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x 7) By applying division algorithm prove that the polynomial g(x) = x2 + 3x + 1 is a factor of the
polynomial f(x) = 3x4 + 5x3 – 7x2 + 2x + 2.
8) What must be subtracted from 8x4 + 14x3 – 2x2 + 7x – 8 so that the resulting polynomial is exactly divisible by 4x2 + 3x – 2.
9) Find the values of a and b so that x4 + x3 + 8x2 + ax + b is divisible by x2 + 1.
24 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume 10) A polynomial f(x) = x4 – 3x3 + 6x – 4 is factorized into three polynomials such that
f(x) = p(x).q(x).r(x). If p(x) = x2 – 3x + 2 and q(x) = x – 2, then find r(x).
11) For which value of a, (x + a) is a factor of 2x2 + 2ax + 5x + 10?
HW Exercise:
1) Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in p(x) = x4 – 5x + 6, g(x) = 2 – x2.
2) Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1.
3) Obtain all other zeroes of x4 + x3 – 34x2 – 4x + 120, if two of its zeroes 2 and – 2. [CBSE–08]
4) On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4, respectively. Find g(x).
5) If (x – 2) is a factor of polynomial x3 + ax2 + bx + 16 and a – b = 6 then find the value of a and b.
6) Apply the division algorithm to find the quotient remainder on dividing f(x) by g(x) as given below.
i) f(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2 ii) f(x) = x4 – 5x + 6, g(x) = 2 – x2
7) What must be added to f(x) = 4x4 + 2x3 – 2x2 + x – 1 so that the resulting polynomial is divisible by g(x) = x2 + 2x – 3.
8) If the polynomial f(x) = 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.
MISCELLANEOUS EXERCISE:
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;
2
1, 1, – 2 ii) x3 – 4x2 + 5x – 2; 2, 1, 1
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.
3) If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
4) If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± 3 find other zeroes.
5) If the polynomial 6x4 + 8x3 + 17x2 + 21x + 7 is divided by another polynomial 3x2 + 4x + 1, the
remainder comes out to be (ax + b), find a and b. [CBSE 09]
6) Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:
i) f(x) = x2 – 2x – 8 ii) g(s) = 4s2 – 4s + 1 iii) h(t) = t2 – 15 iv) p(x) = x2 + 2 2x – 6 v) q(x) = 3 x2 + 10x + 7 3 vi) f(x) = x2 – ( 3 + 1)x + 3 vii) g(x) = a(x2 + 1) – x(a2 + 1)
7) If α and β are the zeros of the quadratic polynomials f(x) = ax2 + bx + c, then evaluate:
i) α – β ii)
−β α
1
1 iii) − αβ
+β
α1 1 2 iv) α2β + αβ2 8) If α and β are the zeros of the quadratic polynomial f(x) = 6x2 + x – 2, find the value of
α +β β α .
9) If α and β are the zeros of the quadratic polynomial f(x) = x2 + x – 2, find the value of
−β α
1 1 .
10) If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of +β
α 1
1 – 2αβ.
Chapter 02: Polynomials 25
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 25 11) If α and β are the zeros of the quadratic polynomial f(t) = t2 – 4t + 3, find the value of α4β3 + α3β4. 12) If α and β are the zeros of the quadratic polynomial p(y) = 5y2 – 7y + 1, find the value of
+β α
1 1 . 13) If the sum of the zeros of the quadratic polynomial f(t) = kt2 + 2t + 3k is equal to their product,
find the value of k.
14) If one zero of the quadratic polynomial f(x) = 4x2 – 5kx – 9 is negative of the other, find the value of k.
15) If α and β are the zeros of the quadratic polynomial f(x) = x2 – 1, find a quadratic polynomial whose zeros are
β α 2 and
α β 2 .
16) Verify that the numbers given along side of the cubic polynomials below are their zeros. Also, verify the relationship between the zeros and coefficients in each case:
i) f(x) = 2x3 + x2 – 5x + 2; ½, 1, –2 ii) g(x) = x3 – 4x2 + 5x – 2; 2, 1, 1
17) Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time, and product of its zeros as 3, –1 and –3 respectively.
18) Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:
i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x2 + x + 1
ii) f(x) = 10x4 + 17x3 – 62x2 + 30x – 3, g(x) = 2x2 + 7x + 1 iii) f(x) = 4x4 + 8x + 8x2 + 7, g(x) = 2x2 – x + 1
iv) f(x) = 15x3 – 20x2 + 13x – 12, g(x) = 2 – 2x + x2
19) Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
i) g(t) = t2 – 3, f(t) = 2t4 + 3t3 – 2t2 – 9t – 12 ii) g(x) = x3 – 3x + 1, f(x) = x5 – 4x3 + x2 + 3x + 1 iii) g(x) = 2x2 – x + 3, f(x) = 6x5 – x4 + 4x3 – 5x2 – x – 15
20) Obtain all zeros of the polynomial f(x) = 2x4 + x3 – 14x2 – 19x – 6, if two of its zeros are –2 & –1.
21) Obtain all zeros of f(x) = x3 + 13x2 + 32x + 20, if one of its zeros is –2.
22) Obtain all zeros of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6, if two of its zeros are – 3 & 3 . 23) Find all zeros of the polynomial f(x) = 2x4 –2x3 – 7x2 + 3x + 6, if two of its zeros are –
2 3 &
2 3 . 24) What must be added to the polynomial f(x) = x4 + 2x3 – 2x2 + x – 1 so that the resulting
polynomial is exactly divisible by x2 + 2x – 3?
25) What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21 so that the resulting polynomial is exactly divisible by x2 – 4x + 3?
26) Find all the zeroes of the polynomial x3 + 3x2 – 2x – 6, if two of its zeroes are – 2 and 2.
[CBSE 09]
27) If the polynomial 6x4 + 8x3 – 5x2 + ax + b is exactly divisible by the polynomial 2x2 – 5, then find
the values of a and b. [CBSE 09]
28) If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is reciprocal of the other, find the value of a.
[CBSE 08]
29) If the product of zeroes of the polynomial ax2 – 6x – 6 is 4, find the value of ‘a’. [CBSE 08]
30) Find the condition which must be satisfied by the coefficient of the polynomial f(x) = x5 – px2 + qx – r when the sum of its two zeros is zero.
31) Find the value of constant k if two of the zeros of the polynomial f(x) = x3 – 3x2 – 4x + k, are equal in magnitude but opposite in sign.
32) Find the zeros of the polynomial f(x) = x3 – 12x2 + 39x – 28, if it is given that the zeros are in A.P.
26 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume 33) Find the value of p if three consecutive odd integers a, b and c are zeros of the polynomial
f(x) = x3 – 15x2 + 71x + p.
34) Find the condition that the zeros of the polynomial f(x) = x3 – px2 + qx – r may be in arithmetic progression.
35) Find the zeros of the polynomial f(x) = x3 – 5x2 – 2x + 24, if it is given that the product of its two zeros is 12.
36) If the zeros of the polynomial f(x) = x3 – 3x2 + x + 1 are a– b, a, a + b, find a and b.
MULTIPLE CHOICE QUESTIONS:
CW Exercise:
1) If α, β are the zeros of the polynomial f(x) = x2 + x + 1, then α
1 + β 1 =
a) 1 b) –1 c) 0 d) None of these
2) If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x + 4k is reciprocal of the other, then k =
a) 2 b) –2 c) 1 d) –1
3) If α and β are the zeros of the polynomial f(x) = x2 + px + q, then a polynomial having α 1 and β
1 is its zeros is
a) x2 + qx + p b) x2 – px + q c) qx2 + px + 1 d) px2 + qx + 1
4) If α, β are the zeros of the polynomial f(x) = x2 – p (x + 1) – c such that (α + 1) (β + 1) = 0, then c =
a) 1 b) 0 c) –1 d) 2
5) If the product of zeros of the polynomial f(x) = ax3 – 6x2 + 11x – 6 is 4, then a =
a)
2
3 b) –
2
3 c)
3
2 d) –
3 2
6) If one root of the polynomial f(x) = 5x2 + 13x + k is reciprocal of the other, then the value of k is
a) 0 b) 5 c)
6
1 d) 6
7) If x = 2 and x = 3 are zeros of the quadratic polynomial x2 + ax + b, the values of a and b respectively are
a) 5, 6 b) –5, –6 c) –5, 6 d) 5, 6.
8) If f(x) = 4x3 – 6x2 + 5x – 1 and α, β and γ are its zeros then αβγ is equal to:
a)
2
3 b)
4
5 c) –
2
3 d)
4 1
9) On dividing x3 – 3x2 + x + 2 by polynomial g(x), the quotient and remainder were x – 2 and 4 – 2x respectively then g(x)
a) x2 + x + 1 b) x2 + x – 1 c) x2 – x – 1 d) x2 – x + 1.
10) If sum of zeros = 2, product of its zeros = 3
1. The quadratic polynomial is
a) 3x2 – 3 2x + 1 b) 2x2 + 3x + 1 c) 3x2 – 2 3 x + 1 d) 2x2 + x + 3 11) Let p(x) = ax2 + bx + c be a quadratic polynomial can have at most
a) one zero b) two zeros c) three zeros d) none of these.
Chapter 02: Polynomials 27
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 27 HW Exercise:
1) If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7, then α 1 +
β
1 is equal to
a)
3
7 b) –
3
7 c)
7
3 d) –
7 3
2) If the sum of the zeros of the polynomial f(x) = 2x3 – 3kx2 + 4x – 5 is 6, then the value of k is
a) 2 b) 4 c) –2 d) –4
3) If α, β are the zeros of polynomial f(x) = x2 – p (x + 1) – c, then (α + 1) (β + 1) =
a) c – 1 b) 1 – c c) c d) 1 + c
4) If the product of two zeros of the polynomial f(x) = 2x3 + 6x2 – 4x + 9 is 3, then its third zero is
a)
2
3 b) –
2
3 c)
2
9 d) –
2 9
5) If the polynomial f(x) = ax3 + bx – c is divisible by the polynomial g(x) = x2 + bx + c, then ab =
a) 1 b)
c
1 c) –1 d) –
c 1
6) If 3 is a zero of the polynomial f(x) = x4 – x3 – 8x2 + kx + 12, then the value of k is:
a) –2 b) 2 c) –3 d)
2 3
7) The sum and product zeros of the quadratic polynomial are –5 and 3 respectively the quadratic polynomial is equal to:
a) x2 + 2x + 3 b) x2 – 5x + 3 c) x2 + 5x + 3 d) x2 + 3x – 5.
8) If the polynomial 3x2 – x3 – 3x + 5 is divided by another polynomial x – 1 – x2, the remainder comes out to be 3, then quotient polynomials is
a) 2 – x b) 2x – 1 c) 3x + 4 d) x – 2.
9) If α, β and γ are the zeros of the cubic polynomial such that α + β + γ = 2, αβ + βγ + γα = –7, αβγ = –14 then cubic polynomial is:
a) x3 – 7x2 – 2x + 14 b) x3 + 2x2 + 7x – 14 c) x3 – 2x2 + 7x + 14 d) x3 – 2x2 – 7x + 14.
10) If the sum of zeros of the polynomial p(x) = kx3 – 5x2 – 11x – 3 is 2 then k is equal to:
a) k = – 2
5 b) k =
5
2 c) k = 10 d)
2 5
11) If 2 and – 2
1 as the sum and product of its zeros respectively then the quadratic polynomial f(x) is:
a) x2 – 2x – 4 b) 4x2 – 2x + 1 c) 2x2 + 4x – 1 d) 2x2 – 4x – 1.
COLUMN MATCHING QUESTIONS:
1) Listed in column I are some types of Polynomials. Choose all correct options for each item in column I in column II.
Column I Column II i) Linear A) 3x2 + 1 ii) Binomial B) x3 – 2 iii) Quadratic C) x3 – 5x + 1 iv) Trinomial D) 2x –1
E) x2 – 5x + 6
28 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume 2) Given in column I are the number of zeros of the Polynomials and column II shows the graphs of
the Polynomial. Choose all correct options for each item in column I in column II.
Column I Column II
i) 2 A)
ii) 1 B)
iii) 0 C)
iv) 3 D)
E)
3) Listed in column I are the zeros of the polynomials and in column II some polynomials. Choose all correct options for each item in column I in column II.
Column I Column II i) –1 and 2 A) x2 – x + 2 ii) –2 and 1 B) x2 + 3x + 2 iii) –2 and –1 C) x2 + x + 2 iv) No zeros D) x2 – x – 2
E) x2 + x – 2
4) In column I are the sum and product of the zeroes of the polynomials. For each item in column I.
Choose all correct options in column II.
Column I Column II
i) S = 3/2, P = –9/2 A) x2 + x – 6 ii) S = –1, P = –6 B) x2 + 3x + 2 iii) S = –3, P = 2 C) 3x2 + 2x + 2
D) 2x2 + 2x – 12
E) 2x2 – 9 – 3x
Y
Y’
X’ 0 X
Y
Y’
X’ 0 X
Y
Y’
X’ 0 X
Y
Y’
X’ 0 X
Y
Y’
X’ 0 X
Chapter 02: Polynomials 29
Volume Universal Tutorials – X CBSE (2012–13) – Mathematics 29 5) Given in column I are some polynomials with one of the zeroes given in brackets. Choose one
correct option from column II for each item in column I.
Column I Column II
i) x2 – x – (2k + 2), (–4) The value of k is
A) 1 ii) x2 – 2x – (7p + 3), (–4)
The value of k is
B) –5 iii) kx2 – 3(k – 1)x – 1, (1)
The value of k is
C) 9 iv) x3 – 4x2 + kx – 2, (2)
The value of k is
D) 3
E) 5
6) Given in column I are some polynomials and column II shows the zeroes of the polynomial.
Choose one correct option from column II for each item in column I.
Column I Column II i) 3x2 + 11x – 4 A) 4, –1 ii) x2 – 16 B) –4, 2 iii) x2 – 3x – 4 C) 4, –4 iv) x2 + 5x + 4 D) –4, 1/3
E) –1, –4
ANSWER TO UNSOLVED EXERCISE:
CW Exercise 2.1:
1) (i) No (ii) 1 (iii) 3 (iv) 2 (v) 4 (vi) 3 CW Exercise 2.2:
1) (i) –2, 4 (ii) ½, ½ (iii) –1, 4/3 iv) 7, –7 2) (i) 3x2 –3 2x + 1 (ii) 4x2 +x + 1 (iii) x2 –4x + 1 3) α =
4 3 , β =
3
−2 4) x = –b/a, c/b 5) (i) p2 – 2q, (ii) p/q
6) (i) 2 22 a
ac b − (ii)
ac ac
b2−2 (iii) 3 3 3
a b
abc− (iv) 3 3 3
c b abc− (v)
c a
b abc
2
3 − 3
7) (i) 4
2 2 2
2 2 ) 2
(
a
c a ac
b − − (ii) ( 2 2 2)22 2 2 2 c
a
c a ac
b − − 8) k = –1 or 2/3 9) x2 – 4x – 5
HW Exercise 2.2:
1) (i) –1/3, 3/2 (ii) –2, 0 (iii) – 15 , 15 2) (i) 4x2 – x – 4 (ii) x2 + 5 (iii) x2 – x + 1
3) 15/4 4) –5/16 5) 6 6) 2
7) 12 8) x2 – x + 41
CW Exercise 2.3:
1) (i) Q = x – 3; R = 7x – 9 (ii) Q = x2 + x – 3; R = 8
2) (i and ii) Yes 3) –1, 1 4) x2 + 2x +1 5) i) P(x) = 2x2 – 2x + 14; g(x) = 2; q(x) = x2 – x + 7; r(x) = 0
2 25
30 Universal Tutorials – X CBSE (2012–13) – Mathematics Volume ii) P(x) = x3 + x2 + x + 1; g(x) = x2 – 1; q(x) = x + 1; r(x) = 2x + 2
iii) P(x) = x3 + 2x2 – x + 2; g(x) = x2 – 1; q(x) = x + 2; r(x) = 4
6) i) Q = x2 – 8x + 27, R = –60 ii) Q = x2 + x – 3, R = 8 8) 14x – 10 9) a = 1, b = 7 10) (x + 2) 11) 2 HW Exercise 2.3:
1) i) Q = –x2 – 2; R = –5x + 10 2) No 3) 5, –6
4) x2 – x + 1 5) a = –2, b = –8 6) i) Q = x–3, R = 7x–9 (ii) Q = –x2–2, R=–5x+10 7) 61x – 65 8) k = 5, a = –5
Miscellaneous:
2) x3 – 2x2 – 7x + 14 3) a = 1; b = ± 2 4) –5, 7 5) k = 5 and a = –5
7) i) a
ac
b2−4 ii) c
ac
b2−4 iii) ⎟
⎠
⎜ ⎞
⎝⎛ +
− a
c c b 2
iv) 2
a
−bc
8) 12
−25 9)
2
−3 10)
4
−27 11) 108
12) 7 13) –2/3 14) 0 15) f(x) = k(x2 + 4x + 4)
17) f(x) = k(x3 – 3x2 – x + 3) 20) –
2
1, 3, –2, –1 21) –10, –1, –2 22) – 3 , 3 , 1, 2
23) 2, –1, 2 3 , –
2
3 24) x – 2 25) 2x – 3 26) –3
28) 3 29) –
2
3 30) pq = r 31) 12
32) 1, 4, 7 33) –105 34) 2p3 – 9pq + 27r = 0 35) 3, 4, –2 36) a = 1, b =
Column Matching Question:
1) i–D; ii–ABD; iii–AE; iv–CE 2) i–C; ii–DE; iii–B; iv–A 3) i–D; ii–E; iii–B; iv–AC 4) i–E; ii–AD; iii–B 5) i–C; ii–D; iii–A; iv–E 6) i–D; ii–C; iii–A; iv–E
± 2