Polynomials Grade 10

[Pages:13]Choose correct answer(s) from the given choices

ID : ca-10-Polynom ials [1]

Grade 10 Polynomials

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(1)

Find

the

zeros

of

the

polynomial

f (x )

=

3

x

-

2

5x

-

36x

+

180,

if

it

is

given

that

two

of

its

zeros

are equal in magnitude but opposite in sign.

a. 5, 8 and -6

b. 6, -6 and 5

c. 6, -6 and -5

d. 5, -3 and 6

(2)

If

the

zeros

of

the

polynomial

3

x

+

2

8x

-

80x

-

384

are

(a

-

b),

a,

(a

+

b),

find

a

and

b.

-8

304

a. a =

and b = ?

3

3

-8

304

b. a =

and b = ?

3

2

3

304

c. a =

and b = ?

-8

3

d. None of these

(3) Find the zeros of the polynomial f(x) = x3 - 18x2 + 104x - 192, if it is given that the zeros are in arithmetic progression.

a. 5, 7 and 9

b. 4, 6 and 8

c. 6, 8 and 10

d. 3, 5 and 7

(4)

Find

the

zeros

of

the

polynomial

2

45p

+

93p

+

42.

6

5

a. p = -

or -

9

7

9

7

b. p = -

or -

6

5

9

5

c. p = -

or -

6

7

6

7

d. p = -

or -

9

5

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ID : ca-10-Polynom ials [2]

(5) The function f is defined by a polynomial. Some values of x and f(x) are shown in the table above. Which of the following must be a factor of f(x)?

x

f (x )

5

0

-6

-6

-5

5

-7

-3

a. x - 5 c. x - 3

b. x - 7 d. x - 8

(6)

Nancy

tried

hard

to

solve

the

quadratic

equation

2

-3z

+

5z

+

3

=

0

but

she

failed

to

do

the

task.

She did the calculation repeatedly, but the answers obtained each time were different. Then what is

true?

a. The equation has two real solutions c. The equation has no real solution

b. The equation has a real and an imaginary solution

d. The equation has only a single solution

(7) If and are the zeros of polynomial x2 - 10x + k, such that - = 2. Find the value of k.

a. 30

b. 28

c. 18

d. 24

Answer the questions

(8)

Find

the

zeros

of

the

polynomial

f (x )

=

3

x

-

2

11x

+

38x

+

40,

if

it

is

given

that

product

of

its

two zeros is 8.

(9)

It

is

given

that

-7

is

a

zero

of

the

polynomial

3

(x

-

2

x

-

44x

+

84),

find

its

other

zeros.

(10)

Find

the

zeros

of

the

polynomial

f(x)

=

3

x

-

2

9x

+ 26x - 24, if it is given that sum of its two

zeros is 7.

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ID : ca-10-Polynom ials [3]

Grade 10 Polynomials

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(1) b. 6, -6 and 5

Step 1

On

comparing

the

polynomial

3

x

-

2

5x

-

36x

+

180,

with

the

standard

form

3

ax

+

2

bx

+

cx

+

d

=

0,

we

get:

a=1 b = -5 c = -36 d = 180

Step 2 Let , and be the zeros of polynomial.

Step 3 It is given than = -

Step 4 Sum of zeros,

-b ++ =

a 5 ++ = 1 + (-) + = 5

=5

Step 5

Product of zeros,

d =

a 180 =

1 ()(-)(5) = 180

180

2

= -5

2

= -36

= 6 or - 6

Step 6 Now, = -6 or 6

Step 7 Hence, zeros are 6, -6 and 5.

ID : ca-10-Polynom ials [4]

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(2) a.

-8

304

a=

and b = ?

3

3

ID : ca-10-Polynom ials [5]

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

-

80x

-

384

with

the

polynomial

given

above,

we

have

+ + = -8 + + = -80

= 384

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

-8 + + = -8 (a - b) + a + (a + b) = -8 3a = -8 a =

3

+ + = -80 a(a - b) + a(a + b) + (a + b)(a - b) = -80

2

2

3a - b = -80

-8

2

2

3(

) - b = -80

3

64

64

304

2

2

2

- b = -80 b =

+ 80 = b =

3

3

3

304

b = ?

3

-8

304

a=

and b = ?

3

3

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(3) b. 4, 6 and 8

Step 1 Let = (a - d), = a and = (a + d) be the zeros of polynomial Step 2 + + = 18 Step 3 (a - d) + a + (a + d) = 18 Step 4 3a = 18 Step 5 a = 18/3 = 6 Step 6 Product of zeros = = 192 Step 7 (a - d)(a)(a + d) = 192 Step 8 a(a2 - d2) = 192 Step 9 6(62 - d2) = 192 Step 10 62 - d2 = 192/6 = 32 Step 11 d2 = 4 Step 12 d = 2 or - 2 Step 13 Hence zeros are (6 - 2), 6 and (6 + 2) => 4, 6 and 8 OR (6 + 2), 6 and (6 - 2) => 8, 6 and 4

ID : ca-10-Polynom ials [6]

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6

7

(4) d. p = - or -

9

5

Step 1 Let us factorize the polynomial by splitting the middle term.

2

f(x) = 45p + 93p + 42

2

= 45p + 63p + 30p + 42 = 9p(5p + 7) + 6(5p + 7) = (9p + 6)(5p + 7)

Step 2 We know that a real number is called a zero polynomial f(x), if f() = 0.

f(x) = 0 (9p + 6)(5p + 7) = 0

(9p + 6) = 0 or (5p + 7) = 0

6

7

p = - or p = -

9

5

6

7

So, the zeros of f(x) are - and - .

9

5

ID : ca-10-Polynom ials [7]

(5) a. x - 5

Step 1 According to the factor theorem, if f(x) is a polynomial of degree greater than or equal to 1 and a is any real number, then x - a is a factor of f(x) if f(a) = 0. Step 2 Now, from the table given in the question, we have f(x) = 0 when x = 5. x - 5 is a factor of f(x). Step 3 Hence, option a is the correct answer.

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(6) a. The equation has two real solutions

ID : ca-10-Polynom ials [8]

Step 1

Here, we have to find the nature of the roots of a quadratic equation and consequently predict

the solution. Every quadratic equation has degree 2 and as a result, the number of roots of a

quadratic

equation

with

the

general

form

2

ax

+

bx

+

c

=

0

is

2.

Step 2

The

roots

of

such

an

expression

of

the

form

2

az

+

bz

+

c

=

0,

where

a,

b

and

c

are

-b ? D

constants are given by z =

. The nature of roots is determined by the value of

2a

the

discriminant,

D

which

is

equal

to

2

b

-

4ac.

1. If D > 0, the roots are real and unequal. 2. If D = 0, the roots are real and equal. 3. If D < 0, the roots are unequal and imaginary.

Step 3

On comparing the given equation with the general equation,

a = -3, b = 5, c = 3

So,

the

value

of

D

=

2

(5)

-

4

?

(-3)

?

(3)

=

61

Step 4

As can be observed from the results mentioned above, the value of D is Positive, hence the roots of the given equation are real and unequal i.e. the equation will have two real roots. Hence, option a matches with the result.

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