RS Aggarwal Solutions Class 10 Ex 2A

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

In Δ ABC, a line DE is drawn joining the midpoints of AB and BC. Which of the following statements is true?