Polynomials Grade 10 - Edugain Math

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Grade 10 Polynomials

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Choose correct answer(s) from the given choices

(1) Find the sum of the zeroes of the polynomial q2 + q - 12.

a. 2

b. 1

c. -2

d. -1

(2)

If

the

zeros

of

the

polynomial

3

x

-

2

11x

+

38x

-

40

are

(a

-

b),

a,

(a

+

b),

find

a

and

b.

11

7

a. a =

and b = ?

3

3

11

7

b. a =

and b = ?

3

2

3

7

c. a =

and b = ?

11

3

d. None of these

(3) Find the product of the zeroes of the polynomial x2 + 4x - 32.

a. 32

b. -16

c. -64

d. -32

(4) If and are the zeros of polynomial x2 - 6x + k, such that - = 4. Find the value of k.

a. 5

b. 4

c. 6

d. 0

Answer the questions

(5) If two zeros of polynomial x3 + bx2 + cx + d are 4+3 and 4-3, find its third zero. (6) If and are the zeros of quadratic polynomial x2 - 5x + 1, find the value of 3 + 3. (7) Find a quadratic polynomial, the sum and product of whose zeros are 4 and -5 respectively. (8) If and are the zeros of polynomial x2 - 4x + k, such that 2 + 2 = 10. Find the value of k.

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ID : w w -10-Polynom ials [2]

(9) For a positive integer k, we write

where 2

k

(1 + x)(1 + 2x)(1 + 3x) ... (1 + kx) = a0 + a1 x + a2 x + ... + akx ,

a0, a1, ... ak are the coefficients of the polynomial. Find the smallest possible value of k if

a0 + a1 + a2 + ... + ak-1 is divisible by 4763.

(10) Find a quadratic polynomial whose zeros are reciprocals of the zeros of the polynomial . 2

x + 5x + 6

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Solutions

(1) d. -1

ID : w w -10-Polynom ials [3]

Step 1

We know that the sum of the zeroes, or the roots of any 2nd degree polynomial of the form aq2 + bq

-b + c is equal to .

a

Step 2

Comparing the given polynomial with the standard form ax2 + bx + c. We have, b = 1, and a = 1.

-b

- 1

Therefore, the value of

in this case will be

, which is equal to -1.

a

1

Step 3 Hence, the sum of the zeroes of the given polynomial is equal to -1.

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(2) a.

11

7

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

+

38x

-

40

with

the

polynomial

given

above,

we

have

+ + = 11 + + = 38

= 40

ID : w w -10-Polynom ials [4]

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

11 + + = 11 (a - b) + a + (a + b) = 11 3a = 11 a =

3

+ + = 38 a(a - b) + a(a + b) + (a + b)(a - b) = 38

2

2

3a - b = 38

11

2

2

3(

) - b = 38

3

121

121

7

2

2

2

- b = 38 b =

- 38 = b =

3

3

3

7

b = ?

3

11

7

a=

and b = ?

3

3

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(3) d. -32

ID : w w -10-Polynom ials [5]

Step 1

We know that the product of the zeroes, or roots, for a 2nd degree polynomial of the form ax2

c +bx + c is equal to .

a

Step 2

On comparing the given polynomial with the standard form ax2 + bx + c. We have, c = -32,

c

-32

and a = 1. So, the value of will be equal to

, or -32.

a

1

Step 3 Hence, the product of the zeroes for the given polynomial is equal to -32.

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(4) a. 5

ID : w w -10-Polynom ials [6]

Step 1

Comparing the given equation with ax2 + bx + c We have a = 1 , b = -6 and c = k

Step 2

-b We know, Sum of zeros =

a

-(-6)

So, + =

= 6

1

Step 3

c We know, Product of zeros =

a

k So, = = k

1

Step 4 Taking the square of both sides of the equation given in the question, we get : ( - )2 = (4)2

Step 5

Applying the formula (a - b)2 = a2 - 2ab +b2 and adding and subtracting 2 from the LHS obtained, we get : 2 + 2 - 2 + 2 - 2 = 16 Using (a+b)2 = a2 + b2 + 2ab, we get: ( + )2 - 4 = 16

Step 6 Substituting the value of + and in the above equation,we get : (6)2 - 4k = 16

Step 7 Solving the equation, 36 - 4k = 16

Step 8 Hence, 4k = 20

Step 9

20

Therefore, k =

= 5.

4

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(5) -b - 8

Step 1 Let third zero be x Step 2 Sum of zeros = -b/1 = -b Step 3 (4+3) + (4-3) + x = -b Step 4 8 + x = -b Step 5 x = -b - 8

ID : w w -10-Polynom ials [7]

(6) 110

Step 1

Comparing the given equation with ax2 + bx + c We have a = 1, b = -5 and c = 1

Step 2

-b We know, Sum of zeros =

a

-(-5)

So, + =

= 5

1

Step 3

c We know, Product of zeros =

a

1 So, = = 1

1

Step 4

Therefore, 3 + 3 = ( + )3 - 3(+) b3] Using + = 5 and = 1 = (5)3 - 3(1)(5) = 110

[Using the formula: (a+b)3=a3 + 3a2b + 3ab2 +

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(7)

2

x - 4x - 5

ID : w w -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, ( + ) = 4 and = -5

f(x) =

2

x - ( + )x +

2

= x - (4)x - 5

2

= x - 4x - 5

Step 3

Hence,

the

required

polynomial

is

f (x )

=

2

x

-

4x

-

5.

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