PROBLEMS ON EQUALITY OF POLYNOMIALS

Problem 1 :

Find constants a, b and c given that

2x² + 4x + 5 = ax² + [2b - 6]x + c for all x

Solution :

2x² + 4x + 5 = ax² + [2b - 6]x + c

Equating the coefficients of x².

2 = a

Equating the coefficients of x.

4 = 2b - 6

4 + 6 = 2b

10 = 2b

Dividing 2 on each sides.

5 = b

Equating the constants terms.

5 = c

So, the values of a, b and c is 2, 5 and 5.

Problem 2 :

2x³ - x² + 6 = (x - 1)² (2x + a) + bx + c for all x.

Solution :

Given, 2x³ - x² + 6 = (x - 1)² (2x + a) + bx + c

2x³ - x² + 6 = x² + 1 - 2(x)(1) (2x + a) + bx + c

2x³ - x² + 6 = (x² + 1 - 2x) (2x + a) + bx + c

2x³ - x² + 6 = 2x³ + ax² + 2x + a - 4x² - a2x + bx + c

Equating the coefficients of x².

-1 =  a - 4

-1 + 4 = a

3  = a

Equating the coefficients of x.

0 = 2 - 2a + b

0 = 2 - 2(3) + b

0 = 2 - 6 + b

0 = -4 + b

4 = b

Equating the constants terms.

6 = a + c

6 = 3 + c

6 - 3 = c

3 = c

So, the values of a, b and c is 3, 4 and 3.

Find a and b if :

Problem 3 :

z⁴ + 4 = (z² + az + 2) (z² + bz + 2) for all z.

Solution :

z⁴ + 4 = (z² + az + 2) (z² + bz + 2)

Adding and subtracting 4z² on left sides.

z⁴ + 4z² + 4 - 4z² =  (z² + az + 2) (z² + bz + 2)

(z² + 2)² - 4z² =  (z² + az + 2) (z² + bz + 2)

(z² + 2)² - (2z)² =  (z² + az + 2) (z² + bz + 2)

((z² + 2) - 2z) ((z² + 2) + 2z) =  (z² + az + 2) (z² + bz + 2)

(z² + 2 - 2z) (z² + 2 + 2z) =  (z² + az + 2) (z² + bz + 2)

z² + 2 - 2z =  z² + az + 2

Equating the coefficients of z.

-2 = a

z² + 2 + 2z = z² + bz + 2

Equating the coefficients of z.

2 = b

So, the values of a and b is -2 and 2.

Problem 4 :

2z⁴ + 5z³ + 4z² + 7z + 6 = (z² + az + 2) (2z² + bz + 3) for all z.

Solution :

2z⁴ + 5z³ + 4z² + 7z + 6 = (z² + az + 2) (2z² + bz + 3)

(z² + 3z + 2) (2z² - z + 3) = (z² + az + 2) (2z² + bz + 3) 

z² + 3z + 2 = z² + az + 2

Equating the coefficients of z.

3 = a

2z² - z + 3 = 2z² + bz + 3

Equating the coefficients of z.

-1 = b

Problem 5 :

Show that z⁴ + 64 can be factorised into two real quadratic factors of the form z² + az + 8 and z² + bz + 8, but cannot be factorised into two real quadratic factors of the form z² + az + 16 and z² + bz + 4.

Solution :

z⁴ + 64 = (z² + az + 8) (z² + bz + 8)

z⁴ + 16z² + 64 - 16z² =  (z² + az + 8) (z² + bz + 8)

(z² + 8)² - 16z² =  (z² + az + 8) (z² + bz + 8)

(z² + 8)² - (4z)² =  (z² + az + 8) (z² + bz + 8)

(z² - 4z + 8) (z² + 4z + 8) = (z² + az + 8) (z² + bz + 8)

Equating the coefficients of z.

z² - 4z + 8 = z² + az + 8

-4 = a

Equating the coefficients of z.

z² + 4z + 8 = z² + bz + 8

4 = b

Hence, z⁴ + 64 can be factorized into z² + az + 16 and z² + bz + 4.

z⁴+ 64 = (z² + az + 16) (z² + bz + 4)

Hence, z⁴ + 64 can not be factorized into z² + az + 16 and z² + bz + 4. There are no solution.

Problem 6 :

Find real numbers a and b such that

x⁴ - 4x² - 8x - 4 = (x² + ax + 2) (x² + bx - 2)

and hence solve the equation. x⁴ + 8x = 4x² + 4.

Solution :

x⁴ - 4x² - 8x - 4 = (x² + ax + 2) (x² + bx - 2)

Factorise the expression.

x⁴ - 4x² - 8x - 4 = (x² - 2x + 2) (x² + 2x - 2)

(x² - 2x + 2) (x² + 2x - 2) =  (x² + ax + 2) (x² + bx - 2)

(x² - 2x + 2) = (x² + ax + 2)

Equating the coefficients of x.

-2 = a

(x² + 2x - 2) = (x² + bx - 2)

Equating the coefficients of x.

2 = b

So, the values of a and b is -2 and 2.

Given, equation is x⁴ + 8x = 4x² + 4

x⁴ - 4x² + 8x - 4 = 0

(x² + 2x - 2) (x² - 2x + 2 ) = 0

x² + 2x - 2 = 0

a = 1, b = 2 and c = -2

Using quadratic formula.

b²- 4ac = (2)² - 4(1)(-2)

= 4 + 8

= 12