RS Aggarwal Solutions Class 10 Ex 2A

RS Aggarwal Class 10 Ex 2A Chapter 2

Question 1: Find the cubic polynomial whose zeros are 3, ½, -1.   

Solution:

Required polynomial = \((x-3)\left ( x-\frac{1}{2} \right )\left [ (x-(-1)) \right ]=(x-3)\left ( x-\frac{1}{2} \right )(x+1)\)

= \((x-3)(x+1)\left ( x-\frac{1}{2} \right )=(x^{2}-2x-3)\left ( x-\frac{1}{2} \right )\)

= \(\frac{2x^{3}-4x^{2}-6x-x^{2}+2x+3}{2}\) = p(x)

p(x) = \(\frac{2x^{3}-4x^{2}-6x-x^{2}+2x+3}{2}\)

p(x) = \(\frac{1}{2}(2x^{3}-5x^{2}-4x+3)\)

Hence, required polynomial is \((2x^{3}-5x^{2}-4x+3)\)

Question 2: On dividing 4x3 – 8x2 + 8x + 1 by a polynomial g(x), the quotient and the remainder are (2x – 1) and (x + 3) respectively. Find g(x).

Solution:

f(x) = 4x3 – 8x2 + 8x + 1

q(x) = (2x – 1)

r(x) = (x + 3)

By division algorithm, we have

Dividend = (Quotient x Divisor) + Remainder

f(x) = q(x) x g(x) + r(x)

(4x3 – 8x2 + 8x + 1) = (2x – 1) x g(x) + (x + 3)

g(x) = \(\frac{(4x^{3}-8x^{2}+8x+1)-(x+3)}{(2x-1)}\)

= \(\frac{4x^{3}-8x^{2}+7x-2}{2x-1}=2x^{2}-3x+2\)

https://lh3.googleusercontent.com/fLo-MPHv9MyEvIfFoUstuSejyuqpdNJSvURHdJxpI9orI4lk-wTkPHPc0wCbdH7Cp0pP5n6lOFA3tA6d6dWRsGPoGRHbLPHDI56RzDk8nCO4nJV0RlX4bBb8hA3y8vqyWk4AxzqCs-KFbpyhgg

Question 3: Divide 2x2 + x – 15 by x + 3 and verify the division algorithm.

Solution:

The terms of dividend and divisor are in decreasing order

https://lh5.googleusercontent.com/b-vexmi9xfK2PDVL9XXufLriJ1zcSJZMpH4UYtZ2fEL09IwXtNy4ZRre1J_3_GfVy2FHEUioz80SimYN0_Z9gGG4rYDgCldYD5OFi_yIAfaR60h71mgFr06Sp3dPQvPuZef3XyqGCcJ-AoqKkQ

Clearly degree (of remainder) = 0 < degree (x + 3)

Therefore, Quotient = 2x – 5 and remainder = 0

\(\Rightarrow\) (Quotient x divisor) + remainder

= (2x – 5)(x + 3) + 0

= 2x2 + 6x – 5x – 15

= 2x2 + x – 15 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Question 4: Divide -5x2 – 17x + 12 by -5x + 3 and verify the division algorithm.

Solution:

First, we write the terms of dividend and divisor in decreasing order of their degree and then perform the division as shown below.

https://lh3.googleusercontent.com/ESqSYvPl6AA_it8mHu9y454nYdMP6ySybXOVildBnzsjWwwNLSzNvKUHslzpLTkfSptCtqk09oF7GKD-nS2fy_5eYXs35XI-c19PDSVbDEQ81RjbOl5YQtPHnh5DELofhp5dmHVReWDSSYxGNw

Clearly degree (of remainder) = 0 < degree (-5x + 3)

Therefore, Quotient = x + 4 and remainder = 0

\(\Rightarrow\) (Quotient x divisor) + remainder

= (x + 4)(-5x + 3) + 0

= -5x2 + 3x – 20x + 12 = 0

= -5x2 – 17x + 12 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

Question 5: Divide 3x3 – 4x2 + 7x – 2 by x2 – x + 2 and verify the division algorithm.

Solution:

First, we write the given polynomials in standard form in decreasing order of degree and then perform the division as shown below.

https://lh3.googleusercontent.com/1aWMBojYiAo31qHOL5Hi68m2Nf1vclhQN5QzBGDbF2vlnff8fKR-w23ussBSJ1ITLNrOsCjbeLvWTHjYOwen27g214IWi-fsV6xP4BjaMKOXUcxxOIz_iMmhPiGfAcISaWVbG0lVwcp4U5Ep6Q

Clearly degree (of remainder) = 0 < degree (x2 – x + 2)

Therefore, Quotient = (3x – 1), Remainder = 0

\(\Rightarrow\) (Quotient x divisor) + remainder

= (3x – 1)(x2 – x + 2) + 0

= 3x3 – 3x2 + 6x – x2 + x – 2 = 0

= 3x3 – 4x2 + 7x -2 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

Question 6:

Divide -6x3 + x2 + 19x + 6 by –3x2 + 5x + 2 and verify the division algorithm.

Solution:

First, we write the given polynomials in standard form in decreasing order of degree and then perform the division as shown below.

https://lh4.googleusercontent.com/NyPI4iXe_tN5umaSgZm42MDLVh4MXPzyisK9caoxj26b0w5maHgftP0pWcAe7ZRMM9rhiFAph13mgtq1Ih_4fRCwiL2P-wvgxFTqsICU_VHXuale0UXjnLYoWO5BbZ7h0pyPJimlmq2Fm2d5_A

Clearly degree (of remainder) = 0 < degree (-3x2 + 5x + 2)

Therefore, Quotient = (2x + 3), Remainder = 0

\(\Rightarrow\) (Quotient x divisor) + remainder

= (2x + 3)(-3x2 + 5x + 2) + 0

= -6x3 + 10x2 + 4x – 9x2 + 15x + 6

Therefore, -6x3 + x2 + 19x + 6 = dividend

Thus, (Quotient x divisor) + remainder = dividend

Hence, the division algorithm is verified.

Question 7: It being given that 2 is a zero of the polynomial x3 – 4x2 + x + 6, find its other zeros.

Solution:

2 is the zero of the polynomial x3 – 4x2 + x + 6

Therefore, x – 2 is a factor of it

Dividing the given polynomial by x – 2

https://lh5.googleusercontent.com/DokiSix74C4QyqT1lhGQKMTSMDX0IhNfUAMubZFxWLEzi-Tt2iNE3j-bO7uveDvQimSr5MWK2HFElimxgYqu9EaG5XhZAydn_ObldXgLBoM0ecp17eJihh4LSBSlU6ZZzW_Pg0zE0t0aBRn0YQ

Remainder = 0, Quotient of q(x) = x2 – 2x – 3

Now, q(x) = x2 – 2x – 3

= x2 – 3x + x – 3 = x(x – 3) + (x – 3)

= (x – 3)(x + 1)

Other zeros of given cubic polynomial are zeros of q(x)

Therefore, q(x) = 0 \(\Rightarrow\) (x – 3)(x + 1) = 0

\(\Rightarrow\) Either x – 3 = 0 or x + 1 = 0

Therefore, Either x  = 3 or x = -1

Therefore, Other zeros of the given cubic polynomial are 3 and -1

Question 8: It being given that -1 is a zero of the polynomial x3 + 2x2 – 11x – 12, find its other zeros.

Solution:

One zero of the polynomial x3 + 2x2 – 11x – 12 is -1

Therefore, x + 1 is a factor of x3 + 2x2 – 11x – 12

Dividing x3 + 2x2 – 11x – 12 by x + 1

https://lh6.googleusercontent.com/GTzn2UoOTPLLS6tSaKqSukM-11_BH4zppOT2gMeVkmG_Bgt3uKHDYeefLuloa2IYDDfeXicgN5hq27zkwIFPUEIAnX74FTZnzfz_T6Kr2AqWp4Vc_LLe6Q3XXFhXFuBQXodFul2BhF-8zMqZMQ

Quotient q(x) = x2 + x – 12

= x2 + 4x – 3x – 12

= (x + 4)(x – 3)

Other zeros of given polynomial are the zeros of q(x)

Therefore, q(x) = 0

\(\Rightarrow\) (x + 4)(x – 3) = 0

\(\Rightarrow\) Either x + 4 = 0 or x – 3 = 0

\(\Rightarrow\) Either x = -4 or x = 3

Therefore, -4, 3 are the zeros of q(x)

Therefore, The zeros of given polynomial are -4, -1 and 3

Question 9: If 1 and -2 are two zeros of the polynomial (x3 – 4x2 – 7x + 10), find its third zero.

Solution:

1 and -2 are the two zeros of the polynomial

x3 – 4x2 – 7x + 10

Dividing the polynomial by x – 1

https://lh4.googleusercontent.com/L4xeHtpeOvu9TOBIOWgjswKqbk6wVxNb_a4rKD7jSBySewtFCd4Bx0nOjM0VDF5PeXdOzGJxy3OxDYv1fD7fJ0qVf-OdJTIbN0P0prAnUrhNzNYgguUVXUyZGlcR0Ot5w7EahP0S0F7IgQT7dA

Quotient q(x) = x2 – 3x – 10

2 is a zero of given polynomial so it is a zero of x2 – 3x – 10

Dividing x2 – 3x – 10 by x + 2

https://lh5.googleusercontent.com/MDxr4OHkZx4hDQTHCgyigKDY6-lKj7T-dm4LSACErfsV3SSYW0ID6Q_8tVBqWsdgbCdiNcN6HZcHUyYqdrRNOGFYqTslH312kVdj-KPQr4pUQybamHCCOsa7QXnn9h600pkE4Mh5O4W3kVrXog

Third zero of the given polynomial is given by x – 5 = 0 or x = 5

Question 10: If 3 and -3 are two zeros of the polynomial (x4 + x3 – 11x2 – 9x + 18), find all the zeros of the given polynomial.

Solution:

3 and -3 are the two zeros of the polynomial x4 + x3 – 11x2 – 9x + 18

Therefore, (x – 3)(x + 3) = x2 – 9 is a factor of given polynomial

Therefore, Dividing the given polynomial by x2 – 9

https://lh3.googleusercontent.com/HQMlljRIFyP4lPjVOsx0LfSJBpAXzbekTmcn0NxelK53S6crkYBPeq3pJ6aC3u9WrP3ZhkWakrUJ2IFqXFK8h46Tnf5KpFYDHHmfGbfZTkUB1IWzWEm2a7kN8zNR2l5mgogDRUd9UyuG8U5xlg

Therefore, Quotient q(x) = x2 + x – 2

Other zeros of given polynomial are -2 and 1

So zeros of given polynomial are 1, -2,3 and -3

Question 11: If 2 and -2 are two zeros of the polynomial (x4 + x3 – 34x2 – 4x + 120), find all the zeros of the given polynomial.

Solution:

2 and -2 are two polynomials

x4 + x3 – 34x2 – 4x + 120

Therefore, (x – 2)(x + 2) = x2 – 4 will divide the given polynomial completely.

Dividing x4 + x3 – 34x2 – 4x + 120 by x2 – 4

https://lh6.googleusercontent.com/2WB8K33_wqbIA30FMgqf4Xd90QTE7JpovEHTz-GGol07zsLf5cb5nSDqQqu8qg4B4IP_Cp7SPdoN3ERgFIPHYMTGTzWeoSrFnUZc_szT3iwoEdTVHNgiL8xUaoJNt5CMkx0tp8GqpsQJ4luwbQ

Therefore, Quotient q(x) = x2 + x – 30

= x2 + 6x – 5x – 30

= x(x + 6) -5(x + 6)

= (x + 6)(x – 5)

For finding zeros of q(x), q(x) = 0

(x + 6)(x – 5) = 0, x = -6 or 5

Other zeros of given polynomial are -6 and 5

So zeros of given polynomial are 2, -2, -6 and 5

Question 12: Find all the zeros of (x4 + x3 – 23x2 – 3x + 60), if it is given that two of its zeros are \(\sqrt{3}\) and \(-\sqrt{3}\).

Solution:

\(\sqrt{3}\) and \(-\sqrt{3}\) are the zeros of polynomial

x4 + x3 – 23x2 – 3x + 60

Therefore, (x – \(\sqrt{3}\) (x + \(\sqrt{3}\)) = x2 – 3 will divide the given polynomial completely

Dividing x4 + x3 – 23x2 – 3x + 60 by x2 – 3

https://lh4.googleusercontent.com/_Ln5o-lNLXxJluB8-2Ew_NrXQMM9HqAG4iT2T6nGqR35tawWoAc1yAqC4M5bW9o0jWDtWmELe_hA85oEC0tVpnVx-zLcUsPZ2DXQ0mYirxB8oomUkNGb6fiioLRs0lLrWTe1qLcYEj4A2Ev2OQ

Quotient q(x) = x2 + x – 20 = x2 + 5x – 4x – 20

= x(x + 5) -4(x + 5)

= (x + 5)(x – 4)

Other zeros of the given polynomial are the zeros of q(x)

Therefore, q(x) = 0 \(\Rightarrow\) (x + 5)(x – 4) = 0

Or x = -5 or 4

Hence the zeros of given polynomial are \(\sqrt{3}\), \(-\sqrt{3}\), -5, 4

Question 13: Find all the zeros of (2x4 – 3x3 – 5x2 + 9x – 3), it Being given that two of its zeros are \(\sqrt{3}\) and \(-\sqrt{3}\).

Solution:

\(\sqrt{3}\) and \(-\sqrt{3}\) are the zeros of polynomial 2x4 – 3x3 – 5x2 + 9x – 3

Therefore, (x – \(\sqrt{3}\))(x + \(\sqrt{3}\)) = x2 – 3 will divide the given polynomial completely

Dividing 2x4 – 3x3 – 5x2 + 9x – 3 by x2 – 3, we get:

https://lh3.googleusercontent.com/FY742LVVBigKdIPvNqt03SLEC1dHgCNsGB2hskh5J4pSYhvjcuTCAAyzRNApGKeCxf7BM_3eDuGc-hsZLWo0byLRql0uxfP6rdTHoCqP3Ne8lAHuCgawiUlPffEZ-11XF7lMe4uWdNuf5cexxA

Therefore, Quotient q(x) = 2x2 – 3x + 1 = 2x2 -2x –x + 1

= 2x(x – 1) – (x – 1) = (x – 1)(2x – 1)

Other zeros of given polynomial are given by

q(x) = 0 \(\Rightarrow\) (x – 1)(2x – 1) = 0

Therefore, x = 1, ½

Hence, zeros of given polynomial are \(\sqrt{3}\), \(-\sqrt{3}\), 1, ½

Question 14: If two zeros of the polynomial (2x4 – 11x3 + 7x2 + 13x – 7) are (3 + \(\sqrt{2}\)) and (3 – \(\sqrt{2}\)), find other zeros.

Solution:

Sum of 3 + \(\sqrt{2}\) and 3 – \(\sqrt{2}\) = (3 + \(\sqrt{2}\)) + (3 – \(\sqrt{2}\)) = 6

Product of (3 + \(\sqrt{2}\)) and (3 – \(\sqrt{2}\))

= (3 + \(\sqrt{2}\))(3 – \(\sqrt{2}\)) = 9 – 2 = 7

Polynomial whose zeros are 3 + \(\sqrt{2}\) and 3 – \(\sqrt{2}\) is

x2 – (sum of zeros)x + (product of zeros) = x2 – 6x + 7

Dividing p(x) by x2 – 6x + 7

https://lh6.googleusercontent.com/z6FY711mWDz95nNpjtTBzjE2PkdgIVrBXViM8oOCewJJ2xhZXNBPZCo-ew_boYEmAur6K4ZhBNt5ekrn2pejhYtfnn4_kEsJbp824GLqgKeECDVuRpt1oqyVkkzhBU9vf5rg2JVKAXMIjdMPBQ

Quotient = 2x2 + x – 1

Therefore, Other zeros of polynomial p(x) are also the zeros of q(x)

Therefore, q(x) = 2x2 + x – 1 = 2x2 + 2x – x – 1

= 2x(x + 1) – (x + 1) = (x + 1)(2x – 1)

q(x) = 0

\(\Rightarrow\) (x + 1)(2x – 1) = 0

\(\Rightarrow\) Either x + 1 = 0 or 2x – 1 = 0

\(\Rightarrow\) Either x = -1 or x = ½

Therefore, The zeros of given polynomial p(x) are ½, -1, (3 + \(\sqrt{2}\)) and (3 – \(\sqrt{2}\))

Question 15: Obtain all zeros of the polynomial (x4 + 4x3 – 2x2 – 20x – 15), if two of its zeros are \(\sqrt{5}\) and \(-\sqrt{5}\).

Solution:

\(\sqrt{5}\) and \(-\sqrt{5}\) are the zeros of the polynomial

x4 + 4x3 – 2x2 – 20x – 15

Therefore, (x – \(\sqrt{5}\))(x + \(\sqrt{5}\)) = x2 – 5 will divide the given polynomial completely.

Dividing x4 + 4x3 – 2x2 – 20x – 15 by x2 – 5, we get

https://lh6.googleusercontent.com/j17uLfERjkYp2JW8yziWHAPVBm4QXLO6VVzA-d2kEaoXWxfVNgYsXlzmXzE_W6RR_tNWwREtiAFh8d7r2mAhcnk8einn2IQCMGWCuLIDjAZl7Du_1_BvDH0leRvIWFNpvVrTxWWfYZzuG6yqyg

Quotient = x2 + 4x + 3 = x2 + 3x + x + 3

= x(x + 3) + (x + 3) = (x + 3)(x + 1)

Other zeros of the given polynomial are the zeros of q(x)

Therefore, x = -3, -1

Thus, the zeros of the given polynomial are

\(\sqrt{5}\), \(-\sqrt{5}\), -3, -1

Q.16: Find all the zeros of the polynomial \((2x^{4} – 11x^{3} + 7x^{2}+13x – 7)\), it being given that two of its zeros area \((3 + \sqrt{2}) \; and \; (3 – \sqrt{2})\).

Sol:

The given polynomial is \(f(x) = 2x^{4} – 11x^{3} + 7x^{2}+13x – 7\).

Since \((3 + \sqrt{2}) \; and \; (3 – \sqrt{2})\) are the zeros of f(x),

It follows that each one of ( \(x + 3 + \sqrt{2}\) ) and ( \(x + 3 – \sqrt{2}\) ) is a factor of f(x).

Consequently , \(\left [ x – (3 + \sqrt{2} ) \right ] \left [ x – (3 – \sqrt{2} ) \right ] = \left [ (x- 3) – \sqrt{2} \right ] \left [ (x- 3) + \sqrt{2} \right ]\)

= \(\left [ (x – 3)^{2} – 2 ) \right ] = x^{2} – 6x + 7\), which is a factor of f(x)

On dividing f(x) by (\(x^{2} – 6x + 7\)), we get:

https://lh4.googleusercontent.com/kYgVKSB1mjEi0N6WkoQG2xAkc-EqhHMfjSGpBDk_1jhfSj4VFaia8vXl6sphNay0Gvf3dJWPFsrgDXJBYDZXkrYLkqrxzgiPlO7TzBlsbhTCfFnW_5w02AuS_G8TIDmHPICq1O2O

\(Therefore, f(x) = 0\)

\(\Rightarrow 2x^{4} – 11x^{3} + 7x^{2}+13x – 7 = 0 \\ \Rightarrow (x^{2} – 6x + 7) ( 2x^{2} + x – 1) = 0 \\ \Rightarrow (x + 3 + \sqrt{2}) (x+ 3 – \sqrt{2})(2x -1)( x + 1) =0\)

\(\Rightarrow x = -3 – \sqrt{2} \; or \; x = – 3 + \sqrt{2} \; or \; x = \frac{1}{2} \; or \; x = -1\)

Hence, all the zeros are \(\Rightarrow x = \left ( -3 – \sqrt{2} \right ) , \left ( -3 + \sqrt{2} \right ), \frac{1}{2} and -1\)


Practise This Question

CaO(s)+H2O(l)Ca(OH)2(aq)+heat

What can be inferred from the above reaction?

(i)  CaO is solid.

(ii) CaO is reactive with water.

(iii) H2O is liquid.

(iv) H2O is in vapour state.

(v) Ca(OH)2  forms a solution with water

(vi) The reaction is exothermic.