Polynomials.solution

  EXERCISE 2.2

Q.1 Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) 6x² + 11x + 5 = 0

Soln: –

6x2 + 11x + 5 = 6x2 + 6x + 5x + 5

= 6x(x +1) + 5(x +1)

= (x +1) (6x +5)

∴ zeroes of polynomial equation 6x2 +11x +5 are { −1, −56 }

Now, Sum of zeroes of this given polynomial equation = −1+( −56 ) = −116

But, the Sum of zeroes of any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= −116

And Product of these zeroes will be = −1×−56 = 56

But, the Product of zeroes of any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= 56

Hence the relationship is verified.

(ii) 4s2 – 4s + 1

Sol:

     4s2 – 4s + 1 = 4s2 – 2s – 2s + 1

= 2s (2s – 1) –1(2s – 1)

= (2s – 1) (2s – 1)

∴ zeroes of the given polynomial are: {12,12}

∴ Sum of these zeroes will be =  = 1.

But, The Sum of zeroes of any quadratic polynomial equation is given by = −coeff.ofscoeff.ofs2

= −44 = 1

And the Product of these zeroes will be = 12×12
=14

But, Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofs2

= 14.

Hence, the relationship is verified.

(iii) 6x2 – 3 – 7x

Sol:

     6x2 – 7x – 3 = 6x2 – 9x + 2x – 3

= 3x (2x – 3) +1(2x – 3)

= (3x + 1) (2x – 3)

∴ zeroes of the given polynomial are: – (−13,32)

∴ sum of these zeroes will be =  −13+32
=76

But, The Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= 76

And Product of these zeroes will be = −13×32=−12
Also, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= −36 = −12

Hence, the relationship is verified.

(iv) 4u2 + 8u

Sol:

     4u2 + 8u = 4u (u+2)

Clearly, for finding the zeroes of the above quadratic polynomial equation either: – 4u=0 or u+2=0

Hence, the zeroes of the above polynomial equation will be (0, −2)

∴ Sum of these zeroes will be = −2

But, the Sum of the zeroes in any quadratic polynomial equation is given by = −coeff.ofucoeff.ofu2

 = −84 = −2

And product of these zeroes will be = 0 × −2 = 0

But, the product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofu2                                                                          = −04 = 0

Hence, the relationship is verified.

(v) t2 – 15

Sol:

     t2 – 15 = (t+ 15−−√) (t − 15−−√)

Therefore, zeroes of the given polynomial are: – {15−−√, −15−−√}

∴ sum of these zeroes will be = 15−−√ −  15−−√ = 0

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

                                                                = −01 = 0   

And the product of these zeroes will be = (15−−√) × (−15−−√)  = −15

But, the product of zeroes in any quadratic polynomial equation is given by

= constanttermcoeff.oft2                                                                     = −151  = −15

Hence, the relationship is verified.

(vi) 3x2 – x – 4

Sol:

 3x2 − x − 4   = 3x2 – 4x + 3x − 4

= x (3x – 4) +1(3x – 4)

= ( x + 1) (3x – 4)

∴ zeroes of the given polynomial are: – {−1, 43 }

∴ sum of these zeroes will be = −1 +43 = 13

But, the Sum of zeroes in any quadratic polynomial equation is given by = −coeff.ofxcoeff.ofx2

= −(−1)3 = 13

And the Product of these zeroes will be = {−1 × 43 }

= −43

But, the Product of zeroes in any quadratic polynomial equation is given by = constanttermcoeff.ofx2

= −43

Hence, the relationship is verified.

Q2. Form a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

(i). 26 , −3

Sol.

Given,

α + β = 26

αβ = −3

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

Thus, the required quadratic equation will be:

x2 – (26)x −3 = 0

6x2 − 2x – 18 = 0.

(ii). 3–√ , 43

Sol.

Given,

α + β = 3–√

αβ = 43

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (3–√)x + 43 =0

3x2 − 33–√x + 4 = 0.

(iii).  0, 7–√

Sol.

Given,

α + β = 0

αβ = 7–√

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ=0

Thus, the required quadratic equation will be: –

x2 – (0)x + 7–√ = 0

x2 + 7–√ = 0.

(iv).  −2, −2

Sol.

Given,

α + β = −2

αβ = −2

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:

x2 – (α+β)x +αβ=0

∴ The required quadratic polynomial will be:

x2 – (−2)x −2 = 0

x2 + 2x – 2 = 0.

(v). −72, 39

Sol.

Given,

α + β = −72

αβ = 39

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:

x2 – (−72)x + 39 = 0

18x2 + 63x + 6 = 0.

(vi). 6, 0

Sol.

Given,

α + β = 6

αβ = 0

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as:-

x2 – (α+β)x +αβ = 0

∴ The required quadratic polynomial will be:-

x2 – 6x + 0 = 0

x2 – 6x = 0.

                                  EXTRA QUESTIONS

Q.1 Find a quadratic polynomial whose zeroes are: -2+12√, 2−12√.

Sol.

Given: –

α + β = 2+12√ + 2−12√

= 4.

αβ = (2+12√)(2−12√)=4−12=72

∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial equation can be written directly as: –

x2 – (α+β)x +αβ = 0

Thus, the required quadratic polynomial equation will be :-

x2 – (4)x −72 = 0

2x2−8x+7 = 0.

Q.2 If α and β are the roots of a quadratic polynomial ax2+bx+c, then find the value of  α2 + β2.

Sol.

From the equation, (α+β=−ba)

And,               α×β=ca

(α+β)2=α2+β2+(2αβ)

∴ α2+β2=(α+β)2−2αβ

∴ α2+β2=(−ba)2−2ca

∴ α2+β2=b2a2−2ca

 ∴α2+β2=b2−2aca2.

Similarly, we can find out the values of (α3 + β3) and (α3 – β3).

Q3. If α and β are zeroes of a quadratic polynomial x2+4x+3, form the polynomial whose zeroes are  1+αβand1+βα.

Sol.

Since α and β are zeroes of a quadratic polynomial x2+4x+3,

α+β= -4, αβ = 3

Given: –  α1 = 1+αβ

β1 = 1−βα

Now, sum of zeroes = 1+αβ+1+βα

= 2+αβ+βα

= 2αβ+α2+β2αβ

= (α+β)2αβ

On putting values of α+β and αβ from above we get:-

Sum of zeroes = α1 + β1 = −423

= 163

Now, Product of zeroes = (1+αβ)(1+βα)

=1+βα+αβ+αββα

=2αβ+β2+α2αβ

= (α+β)2αβ

On putting values of α+β and αβ we get:

Product of zeroes = α1 × β1 = −423

= 163

Thus the required quadratic polynomial equation will be:-

x2 – (α1 + β1)x + α1β1 = 0

x2 – (163)x + 163 = 0

3x2 – 16x +16=0.

Q4. If α and β are zeroes of a quadratic polynomial p(x) = rx2+4x+4, Find the values of “r” if: – α2 + β2 = 24.

Sol.

From the given polynomial p(x),

α + β = −4r, and αβ = −4r………. (1)

Since,(α + β)2 = α2 + β2 + 2αβ

Given,   α2 + β2 = 24 and from equation (1).

Therefore,  (−4r)2=24+2×4r

16r2=24+2×4r

16 = 24 r2 + 8r

3r2 + r – 2 = 0

3r2 + 3r – 2r -2 = 0

3r(r+1) – 2(r+1) = 0

(3r-2) (r+1) = 0

∴  r = 23   or   r = -1

  EXERCISE 2.3

Q.1 If a polynomial x3 -3x2+x+2 is divided by a polynomial g(x), the quotient and remainder obtained are (x-2) and (-2x+4), respectively. Find the equation of g(x).

Sol.

Since, Dividend = Divisor × Quotient + Remainder

Therefore, x³ -3x2+x+2 = g(x) × (x-2) + (-2x+4)

(x3 -3x2+x+2) – (-2x+4) = g(x) × (x-2)

Therefore, g(x) × (x-2) = x3 -3x2+3x-2

Now, for finding g(x) we will divide “x3 -3x2+3x-2” with (x-2)

Therefore, g(x) = (x2 – x +1)

                                 

Q.2 Find the quotient and remainder by dividing the polynomial f(x) by the polynomial g(x).

(i) f(x) = x3 + 2x2 – 9x + 5,   g(x) = x2+5

Therefore, Quotient is (x+2) and Remainder is (-14x − 5)

(ii) f(x) = x5+2x4-9x3+5x2-2x+1, g(x) = x3 + x2 – x+1

Therefore, Quotient is (x2 + x − 9) and Remainder is (14x2 – 12x +10)

(iii) f(x) = 2x4 + 7x3 + 5x2 + 8x + 5, g(x) = 11− 2x3 + x2

Therefore, Quotient is  – (x + 4) and Remainder is (9x2 + 19x +49).

Q3. Find all the zeroes of the polynomial equation 2x4-3x3-3x2+6x-2, if two of its zeroes are 2–√and−2–√.

Sol.

Since this is a polynomial equation of degree 4, there will be a total of 4 roots.

Let, f(x) = 2x4 – 3x3 – 3x2 + 6x – 2

2–√and−2–√ are zeroes of f(x).

∴(x−2–√)(x+2–√)= (x2− 2) = g(x), is a factor of given polynomial f(x).

If we divide f(x) by g(x), the quotient will also be a factor of f(x) with remainder =0.

So, 2x4 − 3x3 − 3x2 + 6x – 2 = (x2 – 2) (2x2 – 3x +1).

Now, on further factorizing (2x2 – 3x +1) we get,

2x2 – 3x +1 = 2x2 – 2x − x +1 = 0

2x (x − 1) – 1(x−1) = 0

(2x−1) (x−1) = 0

So, its zeroes are given by:  x= 12 and x = 1

Therefore, all four zeroes of the given polynomial equation are:

2–√,−2–√,12and1.