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Exercise 2.4 - Chapter 2 Polynomials class 10 ncert solutions Maths - SaraNextGen [2024]


Question 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:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_e07c2ed.gif

Answer:

(i) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_6067ef5e.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m1ffeec0f.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m3d61ad8a.gif

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m5a4d85ce.gif , 1, and −2 are the zeroes of the given polynomial.

Comparing the given polynomial with https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m3f66d502.gif , we obtain a = 2, b = 1, c = −5, d = 2

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_3a461867.gif

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

(ii) https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m34980791.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_72c249bd.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_30d6a081.gif

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

Comparing the given polynomial with https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m3f66d502.gif , we obtain a = 1, b = −4, c = 5, d = −2.

Verification of the relationship between zeroes and coefficient of the given polynomial

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m50d0d130.gif

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m7afa4813.gif

Multiplication of zeroes = 2 × 1 × 1 = 2 https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1978/Chapter%202_html_m4b6f9650.gif

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

Question 2:

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

Answer:

Let the polynomial be https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1984/Chapter%202_html_m3f66d502.gif and the zeroes be https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1984/Chapter%202_html_m71b5df7c.gif .

It is given that

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1984/Chapter%202_html_m479492dd.gif

If a = 1, then b = −2, c = −7, d = 14

Hence, the polynomial is https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1984/Chapter%202_html_1fae421f.gif .

 

Question 3:

If the zeroes of polynomial https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_794364d0.gif  arehttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_m323d271b.gif , find a and b.

Answer:

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_5ae45263.gif

Zeroes are a − ba + a + b

Comparing the given polynomial with https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_1836d6a7.gif , we obtain

p = 1, q = −3, r = 1, t = 1

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_m5c14590b.gif

The zeroes are https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_m588f29e2.gif .

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_7dcf43eb.gif

Hence, a = 1 and b = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_m74e5c629.gif  or https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1987/Chapter%202_html_m365bb08.gif .

 

Question 4:

]It two zeroes of the polynomial https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_732de33a.gif  arehttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_24202dae.gif , find other zeroes.

Answer:

Given that 2 +https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_14b7969d.gif  and 2­­https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_9189cc1.gif https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_14b7969d.gif  are zeroes of the given polynomial.

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_37f95e5d.gifx2 + 4 ­­− 4x − 3

= x2 ­− 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_732de33a.gif  by x2 ­− 4x + 1.

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_m1c6e559f.gif

Clearly,https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_732de33a.gif  = https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_m36a95250.gif

It can be observed that https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_4b72de00.gif is also a factor of the given polynomial.

And https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_4b72de00.gifhttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_m4eb6becf.gif

Therefore, the value of the polynomial is also zero when https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_m3de706ab.gif or https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1989/Chapter%202_html_44edfe25.gif

Or x = 7 or −5

Hence, 7 and −5 are also zeroes of this polynomial.

 

Question 5:

If the polynomial https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_2f6fd6a.gif  is divided by another polynomialhttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m1e6a6ff6.gif , the remainder comes out to be x + a, find k and a.

Answer:

By division algorithm,

Dividend = Divisor × Quotient + Remainder

Dividend − Remainder = Divisor × Quotient

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m7d0372e1.gif  will be perfectly divisible by https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m1e6a6ff6.gif .

Let us divide https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_4470733.gif  by https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m1e6a6ff6.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_5c6ed6dc.gif

It can be observed thathttps://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m925b52a.gif  will be 0.

Therefore, https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m5acd7939.gif = 0 and https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_6d1224cb.gif = 0

For https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_m5acd7939.gif = 0,

2 k =10

And thus, k = 5

For https://img-nm.mnimgs.com/img/study_content/curr/1/10/9/129/1993/Chapter%202_html_6d1224cb.gif = 0

10 − a − 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0

− 5 − a = 0

Therefore, a = −5

Hence, k = 5 and a = −5

Also Read : Exercise-4.4-Chapter-4-Quadratic-Equations-class-10-ncert-solutions-Maths

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