**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu **10th Maths Book Back Answers Unit 3 – Algebra Ex 3.8** are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete** Samacheer Kalvi Books Back Answers/Solutions** are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.8 Answers in English.

**10th Maths Book Back Answers/Solutions:**

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 8 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

**Chapter 3**

**Exercise 3.8 Algebra**

1.Find the square root of the following polynomials by division method

(i) x^{4} – 12x^{3} + 42x^{2} – 36x + 9

(ii) 37x^{2} – 28x^{3} + 4x^{4} + 42x + 9

(iii) 16x^{4} + 8x^{2} + 1

(iv) 121x^{4} – 198x^{3} – 183x^{2} + 216x + 144

**Solution:**

The long division method in finding the square root of a polynomial is useful when the degrees of a polynomial is higher.

2.Find the square root of the expression x2y2−10xy+27−10yx+y2x2

**Solution:**

3.Find the values of a and b if the following polynomials are perfect squares

(i) 4x^{4} – 12x^{3} + 37x^{2} + bx + a

(ii) ax^{4} + bx^{3} + 361ax^{2} + 220x + 100

**Solution:
**(i)

Since it is a perfect square.

Remainder = 0

⇒ b + 42 = 0, a – 49 = 0

b = -42, a = 49

(ii) ax^{4} + bx^{3} + 361ax^{2} + 220x + 100

Since remainder is 0

a = 144

b = 264

4. Find the values of m and n if the following expressions are perfect squares

(i) 1×4−6×3+13×2+mx+n

(ii) x^{4} – 8x^{3} + mx^{2} + nx + 16

**Solution:**

(i)

(ii)

Since remainder is 0,

m = 24, n = -32

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