# FINDING THE SQUARE ROOT OF A POLYNOMIAL BY LONG DIVISION METHOD

Finding the Square Root of a Polynomial by Long Division Method

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

Here we are going to see how to find square root of a polynomial of degree 4.

## Finding the Square Root of a Polynomial by Long Division Method - Questions

Find the square root of the following polynomials by division method

(i)  x4 −12x3 + 42x2 −36x + 9

Step 1 :

x4 has been decomposed into two equal parts x2 and x2.

Step 2 :

Multiplying the quotient (x2) by 2, so we get 2x2. Now bring down the next two terms -12x3 and 42x2.

By dividing -12x3 by 2x2, we get -6x.  By continuting in this way, we get the following steps. Hence the square root of  x4 −12x3 + 42x2 −36x + 9 is x2 - 6x + 3

(ii) 37x2 −28x3 + 4x4 + 42x + 9

Solution :

First let us arrange the given polynomial from greatest order to least order.

4x4  −28x3 + 37x+ 42x + 9 Hence the square root of  37x2 −28x3 + 4x4 + 42x + 9 is 2x2 - 7x - 3.

(iii) 16x4 + 8x2 + 1

Solution : Hence the square root of  37x2 −28x3 + 4x4 + 42x + 9 is 4x2 + 0x + 1.

(iv) 121x4 − 198x3 − 183x2 + 216x + 144

Solution : Hence the square root of   121x4 − 198x3 − 183x2 + 216x + 144 is 11x2 + 9x + 12.

Question 2 :

Find the square root of the expression

(x2/y2) - 10x/y + 27 - (10y/x) + (y2/x2)

Solution :

By taking L.C.M, we get

(x4 - 10x3y + 27x2y2 - 10xy3+ y4)/x2y2

=  (x4 - 10x3y + 27x2y2 - 10xy3+ y4)/x2y2 =  (x2 - 5xy + y2)/xy

=  (x/y)  - 5 + (y/x)

Hence the square root of the polynomial (x2/y2) - 10x/y + 27 - (10y/x) + (y2/x2) is (x/y)  - 5 + (y/x).

Let us look into the next example on "Finding the Square Root of a Polynomial by Long Division Method".

## How to Find the Missing Values of Square Root of Polynomial

Question 1 :

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

(i) 4x4 −12x3 + 37x2 + bx + a

Solution : By equating the coefficients of x, we get

b = -42

By equating the constant terms, we get

a = 49

Hence the values of a and b are -49 and 42 respectively.

(ii)  ax4 + bx3 + 361x2 + 220x + 100

Solution : Equating the coefficients of x3, we get

b  = 264

By equating the coefficients of x4, we get

a  =  144

Hence the values of a and b are 144 and 264 respectively.

Question 2 :

Find the values of m and n if the following expressions are perfect sqaures

(i)  (1/x4) - (6/x3) + (13/x2) + (m/x) + n

Solution :

By taking L.C.M, we get

(1 - 6x + 13x2 + mx3 + nx4)/x4 By equating the coefficients of x3, we get

m = -12

By equating the coefficients of x4, we get

n = 4

Hence the values of m and n are 6 and 4 respectively.

(ii) x4 − 8x3 + mx2 + nx + 16

Solution : By equating the constant term, we get

[(m - 16)/2]2  =  16

(m - 16)/2  =  4

m - 16  =  8

m  =  8 + 16  =  24

By equating the coefficients of x, we get

n = -4(m - 16)

n = -4(24 - 16)

n = -4(8)  =  -32

Hence the values of m and n are 24 and -32. After having gone through the stuff given above, we hope that the students would have understood, "Finding the Square Root of a Polynomial by Long Division Method".

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