**1. The graphs of y=p(x) are given to us, for some polynomials p(x). Find the number of zeroes of p(x), in each case.**

### (i)

### (ii)

### (iii)

### (iv)

### (v)

### (vi)

### Answer:

### (i) The graph of P(x) does not cut the x-axis at all . So, the number of zeroes is 0.

(ii) The graph of P(x) intersects the x-axis at only 1 point.

So, the number of zeroes is 1.

(iii) The graph of P(x) intersects the x-axis at 3 points.

So, the number of zeroes is 3.

(iv) The graph of P(x) intersects the x-axis at 2 points.

So, the number of zeroes is 2.

(v) The graph of P(x) intersects the x-axis at 4 points.

So, the number of zeroes is 4.

(vi) The graph of P(x) intersects the x-axis at 3 points.

So, the number of zeroes is 3.

# EXERCISE 2.2

**1. Find the zeros of the quadratic polynomials and verify the relationships between the zeros and the coefficients.**

**(i) x ^{2}-2x-8**

**Answer:**

Let, P(x) = x^{2}-2x-8

Comparing P(x) with a quadratic equation ax^{2}+bx+c

we have b = -2 and c = -8.

P(x) = x^{2}-2x-8

= x^{2}-4x+2x-8

= x(x-4) + 2(x-4)

= (x-4) (x+2)

Therefore, the value of P(x) will be zero if x-4=0 or x+2=0 i.e., when x=4 or x=-2

Hence the zeros of P(x) are 4, -2

Sum of zeros = 4+(-2) = 2/1 = â€“-b/a = (-Coefficient of x)/(Coefficient of x2)

Product of the zeros = 4(-2) = -8 = -8/1 = c/a = Constant term / Coefficient of x2

**(ii) 4s ^{2}-4s+1**

**Answer:**

Let P(x) = 4s^{2}-4s+1

To find the zeros of the quadratic polynomial we consider

4s^{2} âˆ’ 4s + 1 = 0

4s^{2} âˆ’ 2s â€“ 2s + 1 = 0

2s(2s â€“ 1) -1(2s â€“ 1) = 0

(2s â€“ 1)(2s â€“ 1) = 0

2s â€“ 1 = 0 and 2s â€“ 1 = 0

2s = 1 and 2s = 1

s = 1/2 and s = 1/2

âˆ´ The zeroes of the polynomial = 1/2 and 1/2

Sum of the zeroes = -(coefficient of s)/(coefficient of s^{2})

Sum of the zeroes = -(-4)/4 = **1**

Letâ€™s find the sum of the roots = 1/2 + 1/2 = **1**

Product of the zeros = Constant term / Coefficient of s^{2}

Product of the zeros =**1 / 4**

Letâ€™s find the products of the roots = 1/2 Ã— 1/2 = **1/4**

**(iii) 6x ^{2}-3-7x**

**Answer:**

Let, P(x) = 6x^{2}-3-7x

**To find the zeros**

Let us put f(x) = 0

â‡’ 6x^{2} â€“ 7x â€“ 3 = 0

â‡’ 6x^{2} â€“ 9x + 2x â€“ 3 = 0

â‡’ 3x(2x â€“ 3) + 1(2x â€“ 3) = 0

â‡’ (2x â€“ 3)(3x + 1) = 0

â‡’ 2x â€“ 3 = 0

**x = 3/2**

â‡’ 3x + 1 = 0

**â‡’ x = -1/3**

It gives us 2 zeros, for x = 3/2 and x = -1/3

Hence, the zeros of the quadratic equation are 3/2 and -1/3.

Now, for verification

Sum of zeros = â€“ coefficient of x / coefficient of x2

3/2 + (-1/3) = â€“ (-7) / 6 7/6 = 7/6

Product of roots = constant / coefficient of x2

3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2

Therefore, the relationship between zeros and their coefficients is verified.

**(iv) 4u ^{2}+8u**

**Answer:**

Let, P(u) = 4u^{2}+8u

Comparing P(u) with a quadratic equation au^{2}+bu+c we have a = 4 ,b= 8 and c = 0

P(u) = 4u^{2}+8u

= u(4u+8)

Therefore, the value of P(u) will be zero

Then, u = 0

Or, 4u+8 = 0

4u = -8

u = -2

Hence the zeros of P(s) are 0, -2

Now sum of zeros = 0+(-2) =-2 = -2

And product of the zeros = 0 X 2= 0

But, the Sum of the zeroes in any quadratic polynomial equation is given by = âˆ’coeff.of u / coeff.of u^{2} = âˆ’8/4 = âˆ’2

and, the product of zeroes in any quadratic polynomial equation is given by = constant term / coeff.of u^{2} = âˆ’0/4 = 0

**(v) t ^{2}-15**

**Answer:**

Let P(t) = t^{2}-15

Comparing P(t) with a quadratic equation at^{2}+bt+c we have a = 1,b= 0 and c = -15

Therefore, the value of P(t) will be zero if t^{2}-15 = 0 i.e., t^{2}= 15 so t = Â±âˆš15

Hence the zeros of P(s) are âˆš15, â€“ âˆš15

Sum of zeros = âˆš15 â€“ âˆš15 = 0 = (âˆ’coeff.of t / coeff.of t^{2})

Product of the zeros = âˆš15(-âˆš15) = -15 =(constant term / coeff.of t^{2})

**(vi) 3x ^{2}â€“ x-4**

**Answer:**

Let P(x) = 3x^{2}-x-4

Comparing P(x) with a quadratic equation ax^{2}+bx+c we have a = 3, b= -1 and c = â€“4

P(x) = 3x^{2}-x-4

= 3x^{2}-4x+3x-4

= x(3x-4)+1(3x-4)

= (3x-4)(x+1)

Therefore the value of P(x) will be zero if 3x-4=0 or x+1=0

Then, 3x = 4

x = 4/3

Or, x = -1

Hence the zeros of P(x) are 4/3, â€“ 1

Sum of zeros = 4/3 â€“ 1= 1/3 = -b/a = (-Coefficient of x)/(Cofficient of x^{2})

Product of the zeros = (4/3) (-1) = â€“4/3 = = (constant term)/(Cofficient of x^{2})

**2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.**

(i) 1/4 , âˆ’1

(ii) âˆš2 , 13

(iii) 0, âˆš5

(iv) 1, 1

(v) -1/4 , 1/4

(vi) 4, 1

**Answer:**

(i) 1/4, -1

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(1/4)x -1 = 0

Multiply by 4 to remove denominator we get

4xÂ² – x -4 = 0

(ii) âˆš2 , 1/3

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(âˆš2)x + 1/3 = 0

Multiply by 3 to remove denominator we get

3xÂ² – 3âˆš2 x + 1 = 0

(iii) 0, âˆš5

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(0)x + âˆš5 = 0

simplify it we get

xÂ² + âˆš5 = 0

(iv) 1, 1

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(1)x + 1 = 0

simplify it we get

xÂ² – x + 1 = 0

(v) -1/4 ,1/4

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(-1/4)x + 1/4 = 0

multiply by 4 we get

4xÂ² + x + 1 = 0

(vi) 4, 1

Now formula of quadratic equation is

xÂ²-(Sum of root)x + (Product of root) = 0

Plug the value in formula we get

xÂ² â€“(4)x + 1 = 0

xÂ² – 4x + 1 = 0

# EXERCISE 2.3

**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) = x^{3} – 3x^{2} + 5x – 3 , g(x) = x^{2} – 2

(ii) p(x) = x^{4} – 3x^{2} + 4x + 5, g(x) = x^{2} – x + 1

(iii) p(x) = x^{4 }– 5x + 6, g(x) = 2 -x^{2}

**Answer:**

(i)p(x) = x^{3} – 3x^{2} + 5x – 3 , g(x) = x^{2} – 2

Therefore, quotient = x â€“ 3 and Remainder = 7x â€“ 9

(ii)

Therefore, quotient = xÂ² + x – 3 and Remainder = 8

(iii)

Therefore, quotient = -x^{2} – 2 and, Remainder = âˆ’5x + 10

**2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.**

(i) t^{2} – 3, 2t^{4} + 3t^{2} – 9t -12

(ii) x^{2} + 3x +1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

(iii) x^{3} – 3x + 1, x^{5} – 4x^{3} + x^{2} + 3x +1

**Answer:**

(i) t^{2} – 3, 2t^{4} + 3t^{2} – 9t -12

Remainder = 0

Hence first polynomial is a factor of second polynomial.

(ii) x^{2} + 3x +1, 3x^{4} + 5x^{3} – 7x^{2} + 2x + 2

Remainder = 0

Hence first polynomial is a factor of second polynomial.

(iii)

Remainder â‰ 0

Hence first polynomial is not factor of second polynomial.

**3. Obtain all other zeroes of ( 3×4 + 6×3 – 2×2 – 10x – 5 ) , if two of its zeroes are âˆš5/3 and -âˆš5/3.**

**Answer:**

Two zeroes of ( 3x^{4} + 6x^{3 }– 2x^{2} – 10x – 5 ) are âˆš5/3 and -âˆš5/3 which means thatis a factor of ( 3x^{4} + 6x^{3 }– 2x^{2} – 10x – 5 ).

Therefore, we divide the given polynomial by xÂ² – 5/3.

we factorize xÂ² + 2x + 1 = (x + 1)Â²

Therefore, its zero is given by x + 1 = 0

â‡’ x = – 1

As it has the term (x + 1)Â²,therefore, there will be 2 zeros at x = – 1.

Hence,the zeroes of the given polynomial âˆš5/3 and -âˆš5/3 are âˆ’1 and âˆ’1.

**4. On dividing ( x ^{3} – 3x^{2} + x + 2 ) by a polynomial g(x), the quotient and remainder were (x-2) and (-2x+4) respectively. Find g(x).**

**Answer:**

Let p(x) = x^{3} – 3x^{2} + x + 2, q(x) = (x â€“ 2) and r(x) = (â€“2x+4)

According to Polynomial Division Algorithm, we have

p(x) = g(x).q(x) + r(x)

â‡’ x^{3} – 3x^{2} + x + 2 =g(x).(xâˆ’2)âˆ’2x+4

â‡’x^{3} – 3x^{2} + x + 2 + 2x âˆ’4 =g(x).(xâˆ’2)

â‡’x^{3} – 3x^{2} + 3x + 2=g(x).(xâˆ’2)

So, Dividing x^{3} – 3x^{2} + 3x + 2 by (xâˆ’2), we get

(x^{2} – x + 1)

Therefore, we haveg(x) = (x^{2} – x + 1)

**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

**Answer:**

Here let us represent, p(x) =dividend=the number to be divided

g(x)=Divisor=the number by which dividend is divided

q(x)=quotient

And r(x)=remainder

Now we have to give examples of polynomials such that the division algorithm

Dividend=Divisor Ã— Quotient +Remainder

Is satisfies and the given conditions are also satisfied

(i)Let us assume the division of 2x + 4 by 2,

Then p(x) = 2x + 4

g(x) = 2

q(x) = x + 2

and r(x) = 0

On using the division algorithm

Dividend=Divisor Ã— Quotient +Remainder

p(x)=g(x) Ã— q(x) +r(x)

On putting the given values we get,

=> 2x + 4 = 2 x (x +2) + 0

On solving we get,

=> 2x + 4 = 2x + 4

Hence the division algorithm is satisfied.

And here the degree of p(x) =1 = degree of q(x)

Hence (i) condition is also satisfied.

(ii) Let us assume the division of x^{3} + x byx^{2}

Then here,p(x) = x^{3}+ x

g(x) = x^{2}

q(x) = x

r(x) = x

It is clear that the degree of q(x) = 1 and

degree of r(x) = 1

on using the devision algorithm

=> Dividend=Divisor Ã— Quotient +Remainder

=> p(x)=g(x) Ã— q(x) +r(x)

On putting the given values we get,

=> x^{3} + x = (x^{2} * x) + x

on solving we get,

x^{3} + x = x ^{3} + x

Hence the division algorithm is satisfied.

and here the degree of r(x) = 1 = degree of q(x)

Hence (ii) condition is also satisfied.

(iii)Let us assume the division of x^{2} + 1 byx

Then here,p(x) = x^{2} + 1

g(x) = x

r(x) = 1

It is clear that the degree of r(x) = 0 and

on using the devision algorithm

=> Dividend=Divisor Ã— Quotient +Remainder

=> p(x)=g(x) Ã— q(x) +r(x)

On putting the given values we get,

=> x^{2} + 1 = (x * x)+ 1

x^{2} + 1 = x^{2} + 1

Hence,the division algorithm is satisfied.

And here the degree of r(x) = 0

Hence (iii) condition is also satisfied.

Here you can also assume any other polynomial for division but it is necessary that the chosen dividend and divisor be such that the conditions of the questions are satisfied.

# Exercise 2.4

**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) 2x^{3} + x^{2} – 5x + 2 ; 1/1,1,-2

#### (ii) x^{3} – 4x^{2} + 5x – 2 ; 2,1,1

**Answer:**

#### (i) 2x^{3 }+ x^{2} â€“ 5x + 2; ^{1}/_{2} , 1, -2

#### p(x) = 2x^{3 }+ x^{2} â€“ 5x + 2

#### Zeros for this polynomial are ^{1}/_{2 }, 1, -2

#### p(^{1}/_{2}) = 2(^{1}/_{2})^{3 }+ (^{1}/_{2})^{2} â€“ 5(^{1}/_{2}) + 2

#### = (^{1}/_{4}) + (^{1}/_{4}) â€“ (^{5}/_{2}) + 2

#### = 0

#### p(1) = 2(1)^{3 }+ (1)^{2} â€“ 5(1) + 2 = 0

#### p(-2) = 2(-2)^{3 }+ (-2)^{2} â€“ 5(-2) + 2

#### = -16 + 4 + 10 + 2

#### = 0

#### Therefore, Â½ , 1, and âˆ’2 are the zeroes of the given polynomial. Comparing the given polynomial with ax^{3} + bx^{2 }+ cx + d, we obtain a = 2, b = 1, c = âˆ’5, d = 2

#### Let us take Î± = ^{1}/_{2} ,Î² = 1, Î³ = -2

#### Î± + Î² + Î³ = ^{1}/_{2} + 1 + (-2) = â€“^{1}/_{2} = â€“^{b}/_{a}

#### Î±Î² + Î²Î³ + Î³Î± = ^{1}/_{2} x 1 + 1(-2) + ^{1}/_{2} (-2) = ^{-5}/_{2 }= ^{c}/_{a}

#### Î±Î²Î³ = ^{1}/_{2} x 1 x (-2) = â€“^{1}/_{1 }= â€“ ^{2}/_{2 }= â€“^{d}/_{a}

#### Therefore, the relationship between the zeroes and the coefficients is verified.

#### (ii) x^{3} â€“ 4x^{2} + 5x â€“ 2; 2,1,1

#### p(x) = x^{3 }â€“ 4x^{2} + 5x â€“ 2

#### Zeros for this polynomial are 2 , 1, 1

#### p(2) = 2^{3 }â€“ 4(2)^{2} + 5(2) â€“ 2

#### = 8 â€“ 16 + (10) â€“ 2

#### = 0

#### p(1) = (1)^{3 }â€“ 4(1)^{2} + 5(1) â€“ 2 = 0

#### Therefore, 2 , 1, and 1 are the zeroes of the given polynomial. Comparing the given polynomial with ax^{3} + bx^{2 }+ cx + d, we obtain a = 1, b = -4, c = 5, d = -2

#### Verify:

#### Sum of zeros = 2+1+1 = 4 = â€“ ^{(-4)}/_{1} = â€“^{b}/_{a}

#### Multiplication of zeroes taking two at a time = (2)(1)+(1)(1)+(2)(1) = 2+1+2= 5 = ^{5}/_{1} = ^{c}/_{a}

#### Multiplication of zeroses = 2 x 1 x 1 = 2 = â€“ ^{(-2)}/_{1} = â€“^{d}/_{a}

#### Hence, the relationship between the zeroes and the coefficients is verified.

**2. Find a cubic polynomial with the sum of the product of its zeroes taken two at a time and the product of its zeroes are 2, -7,-14respectively.**

**Answer:**

#### Let the polynomial be ax^{3} + bx^{2} + cx + d and the zeroes be Î±, Î² and Î³.

#### It is given that

#### Î± + Î² + Î³ = ^{2}/_{1} = â€“^{b}/_{a}

#### Î±Î² + Î²Î³ + Î³Î± = ^{-7}/_{2 }= ^{c}/_{a}

#### Î±Î²Î³ = â€“ ^{14}/_{1 }= â€“^{d}/_{a}

#### If a = 1, then b = âˆ’2, c = âˆ’7, d = 14 Hence, the polynomial is x^{3} â€“ 2x^{2} â€“ 7x + 14

**3. If the zeroes of the polynomialx**^{3} – 3x^{2} + x + 1 are a – b, a , a + b, finda and b .

^{3}– 3x

^{2}+ x + 1 are a – b, a , a + b, finda and b .

**Answer:**

#### p(x) = x^{3} -3x^{2}+x+1

#### Zeroes are a âˆ’ b, a + a + b Comparing the given polynomial with px^{3}+qx^{2}+rx+1, we obtain p = 1, q = âˆ’3, r = 1, t = 1

#### Sum of zeros = a â€“ b + a + a + b

^{-q}/_{ p} = 3a

#### â€“^{(-3)}/_{1 }= 3a

#### 3 = 3a

#### a = 1

#### The zeroes are 1-b, 1, 1+b

#### Multiplication of zeroes = 1(1-b)(1+b)

^{-t}/_{p} = 1 â€“ b^{2}

^{-1}/_{1}= 1 â€“ b^{2}

#### 1 â€“ b^{2} = -1

#### 1 + 1 = b^{2}

#### b = Â±âˆš2

**4. If the two zeroes of the polynomial x**^{4} – 6x^{3} – 26x^{2} + 138x – 35 are 2 + âˆš3, 2 – âˆš3 find other zeroes.

^{4}– 6x

^{3}– 26x

^{2}+ 138x – 35 are 2 + âˆš3, 2 – âˆš3 find other zeroes.

**Answer:**

#### Given 2 + âˆš3 and 2 â€“ âˆš3 are zeroes of the given polynomial. So, (2 + âˆš3)(2 â€“ âˆš3) is a factor of polynomial.

#### Therefore,[x â€“ (2 + âˆš3)][x â€“ (2 â€“ âˆš3)] = x^{2} + 4 âˆ’ 4x âˆ’ 3 = x^{2} âˆ’ 4x + 1 is a factor of the given polynomial

#### For finding the remaining zeroes of the given polynomial, we will find

#### Clearly, x^{4} -6x^{3} -26x^{2}+138x -35 = (x^{2} â€“ 4x + 1)(x^{2} â€“ 2x â€“ 35)

#### We know, (x^{2} â€“ 2x â€“ 35) = (x â€“ 7)(x + 5)

#### Therefore, the value of the polynomial is also zero when (x -7 = 0) or (x + 5 = 0)

#### Or x = 7 or âˆ’5

#### therefore, 7 and âˆ’5 are also zeroes of this polynomial.

**5. If the polynomial x**^{4} – 6x^{3} + 16x^{2} – 25x +10 is divided by another polynomial x^{2} – 2x + k, the remainder comes out to be x+a,findkanda.

^{4}– 6x

^{3}+ 16x

^{2}– 25x +10 is divided by another polynomial x

^{2}– 2x + k, the remainder comes out to be x+a,findkanda.