Polynomials Grade 10

[Pages:9]ID : us-10-Polynom ials [1]

Grade 10 Polynomials

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Choose correct answer(s) from the given choices

1

1

(1)

If

and

are

the

zeros

of

quadratic

polynomial

2

x

+

3x

-

4,

find

the

value

of

+

.

3

3

33

a.

32

67

b.

64

59

c.

64

63

d.

64

(2)

If

the

zeros

of

the

polynomial

3

x

+

2

2x

-

81x

-

162

are

(a

-

b),

a,

(a

+

b),

find

a

and

b.

3

247

a. a =

and b = ?

-2

3

-2

247

b. a =

and b = ?

3

3

-2

247

c. a =

and b = ?

3

2

d. None of these

(3) If two zeros of polynomial x3 + bx2 + cx + d are 2+2 and 2-2, find its third zero.

a. - c - 4

b. b - 4

c. -b - 4

d. c - 4

(4) Find the quadratic polynomial such that sum of its zeros is 17 and difference between zeros is 7.

a. k [x2 - 17x + 60]

b. k [x2 - 17x - 60]

c. k [x2 - 22x + 60]

d. k [x2 + 17x + 60]

Fill in the blanks

1

(5)

If

the

product

of

the

zeros

of

the

polynomial

2

ax

-

16x

+

2

is

, the value of a is

.

15

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ID : us-10-Polynom ials [2]

Answer the questions

11 (6) If a and b are the zeros of quadratic polynomial x2 - 6px + 2q, find the value of + .

a b

(7) If and are the zeros of quadratic polynomial x2 - 3px + q, find the value of 2 + 2.

(8) Find a quadratic polynomial, the sum and product of whose zeros are -9 and -10 respectively.

(9)

If

one

zero

of

the

polynomial

2

(s

+

2

4)z

+

16z

+

4s

is

reciprocal

of

the

other,

find

the

value

of

s.

(10) If and are the zeros of quadratic polynomial x2 + x - 2, find the value of 3 + 3.

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Solutions

63

(1) d.

64

ID : us-10-Polynom ials [3]

Step 1

Comparing

the

given

equation

with

2

ax

+

bx

+

c

We have a = 1, b = 3 and c = -4

Step 2

-b

We know, Sum of zeros =

a

-3

So, + =

= -3

1

Step 3

c

We know, Product of zeros =

a

-4

So, =

= -4

1

Step 4

1

1

Therefore,

+

3

3

3

3

+

= 3 3

3

( + ) - 3( + )

= ()3

3

3

2

2

3

[ Using the formula: (a + b) = a + 3a b + 3ab + b ]

3

(-3) - 3(-4)(-3)

= (-4)3

63 =

64

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(2) b.

-2

247

a=

and b = ?

3

3

Step 1 We know that a cubic polynomial whose zeros are , , and is given by

3

2

p(x) = x - ( + + )x + ( + + )x -

Comparing

3

x

+

2

x

-

81x

-

162

with

the

polynomial

given

above,

we

have

+ + = -2 + + = -81

= 162

ID : us-10-Polynom ials [4]

Step 2 Let . = (a - b), = a, and = (a + b) Thus,

-2 + + = -2 (a - b) + a + (a + b) = -2 3a = -2 a =

3

+ + = -81 a(a - b) + a(a + b) + (a + b)(a - b) = -81

2

2

3a - b = -81

-2

2

2

3(

) - b = -81

3

4

4

247

2

2

2

- b = -81 b =

+ 81 = b =

3

3

3

247 b = ?

3

-2

247

a=

and b = ?

3

3

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(3) c. -b - 4

Step 1 Let third zero be x Step 2 Sum of zeros = -b/1 = -b Step 3 (2+2) + (2-2) + x = -b Step 4 4 + x = -b Step 5 x = -b - 4

(4) a. k [x2 - 17x + 60]

Step 1 Let and be the zeros of the required polynomial. Step 2 Since, sum of zeros + = 17 Step 3 Since, difference of zeros - = 7 Step 4 ( + )2 - ( - )2 = 4 Step 5 (17)2 - (7)2 = 4 Step 6 240 = 4 Step 7 = 240/4 = 60 Step 8 Thus, required polynomial will be k [x2 - ( + )x + ] = k [x2 - 17x + 60]. (where k is a constant.)

ID : us-10-Polynom ials [5]

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(5) 30

Step 1

Let

and

be

the

zeros

of

the

polynomial

2

ax

-

16x

+

2.

constant term

2

Then, =

= .

coefficient of x2

a

1

Also, we are given that = .

15

2

1

Thus, =

. a = 30

a

15

Step 2 Hence, a = 30.

3p (6)

q

Step 1

-6p

Since, sum of zeros = a + b = -

= 6p

1

Step 2

2q

Since, product of zeros = ab =

= 2q

1

Step 3

1 1 a + b 3p

Therefore, + =

=

.

a b

ab

q

ID : us-10-Polynom ials [6]

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(7) 9p2 - 2q

Step 1

Comparing the given equation with ax2+ bx + c We have a = 1, b = -3p and c = q

Step 2

-b We know, Sum of zeros =

a

-(-3p)

So, = + =

= 3p

1

Step 3

c We know, Product of zeros =

a

q So, = = = q

1

Step 4 Applying the formula (a+b)2 = a2 + b2 + 2ab, we get: 2 + 2 = ( + )2 - 2 = (3p)2 - 2(q)

Step 5 Therefore, 2 + 2 = 9p2 - 2q

ID : us-10-Polynom ials [7]

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(8)

2

x + 9x - 10

ID : us-10-Polynom ials [8]

Step 1 We know that when the sum and the product of the zeros of a quadratic polynomial is given, the polynomial can be written as

2

f(x) = x - (Sum of zeros) x + (Product of the zeros)

Step 2 Let and be the zeros of the required polynomial f(x).

Then, ( + ) = - 9 and = -10

f(x) =

2

x - ( + )x +

2

= x - (-9)x - 10

2

= x + 9x - 10

Step 3

Hence,

the

required

polynomial

is

f (x )

=

2

x

+

9x

-

10.

(9) 2

Step 1 Let one zero of the given polynomial be .

1

Then, the other zero is .

1

Product of zeros = ( ? ) = 1.

Step 2

constant term

4s

But, product of zeros =

=

.

coefficient of z 2

(s2 + 4)

4s (s2 + 4)

=1

2

(s + 4) = 4s

2

s + 4 - 4s = 0

2

(s - 2) = 0

s-2= 0 s = 2

Step 3 Thus, the value of s is 2.

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