CHECK WHETHER THE GIVEN POLYNOMIAL IS FACTOR OF ANOTHER POLYNOMIAL USING LONG DIVISION

Question 1 :

Divide the polynomial p(x) by g(x)

p(x) = x³ – 3 x² + 5 x – 3 and g (x) = x² - 2

Solution :

Quotient = x - 3

Remainder = 7 x - 9

Question 2 :

Divide the polynomial p(x) by g(x)

p(x) = x⁴ – 3 x² + 4 x + 5     g (x) = x² + 1 - x

Solution :

Quotient = x² + x - 3

Remainder = 8

Question 3 :

Divide the polynomial p(x) by g(x)

p(x) = x⁴ – 5 x + 6 and g (x) = 2 - x²   

Solution :

Quotient = -x² - 2

Remainder = -5 x + 10

Question 4 :

Check whether the first polynomial is a factor of second polynomial by dividing the second polynomial by the first polynomial

(i) t² – 3 , 2 t⁴ + 3 t³ – 2 t² – 9 t – 12

Since the remainder is 0, we say that the given polynomial is a factor of other polynomial.

(ii)  x² + 3 x + 1     g (x) = 3 x⁴ + 5 x ³ – 7 x² + 2 x + 2     

The remainder is zero. So, we say that the first polynomial is the factor of second polynomial.

(iii)  x³ - 3 x + 1     g (x) = x⁵ -4 x³ + x² + 3 x + 1    

The first polynomial is not a factor of second polynomial.

Question 5 :

Obtain all other zeroes of 3 x⁴ + 6 x³ – 2 x² – 10 x – 5, if two of its zeroes are √(5/3) and -√(5/3)

Solution :

Since the highest power of the given polynomial is 4, there will be four zeroes for the given polynomial. 

Out of the four zeroes, we have two zeroes. By combining these two factors, we will get a quadratic equation. By dividing the given polynomial by this quadratic equation, we will get quotient as quadratic equation.

By solving the quadratic equation that we find quotient, we will get two more zeroes.

x = √(5/3)   x = -√(5/3) 

(x - √(5/3)) (x + √(5/3)) = (x² – 5/3)

3 x² + 6 x + 3  =  3 x² + 3 x + 3 x + 3

                        =  3x (x + 1) + 3 (x + 1)

                        =  (3x + 3) (x + 1)

Question 6 :

On dividing x³ – 3 x² + x + 2 by the polynomial g(x), the quotient and remainder were (x – 2) and – 2x + 4. Find g(x)

Solution :

Division algorithm

P (x) = g (x) x q (x) + r (x)

P (x) = x³ – 3 x² + x + 2

q (x) = (x – 2)

r (x) = - 2x + 4

x³ – 3 x²+ x + 2 = g (x) x (x – 2) + (- 2x + 4)

x³ – 3 x² + x + 2 + 2 x - 4 = g (x) x (x – 2)

x³ – 3 x² + 3 x - 2 = g (x) x (x – 2)

(x³ – 3 x² + 3 x – 2)/(x - 2) = g (x)

Hence, g (x) = x² – 1 x + 1

Question 7 :

Show that (𝑥 + 3) is a factor of 𝑥³ − 19𝑥 − 30. Then find the remaining factors.

Solution :

Let p(x) = 𝑥³ − 19𝑥 − 30

x + 3 = 0

x = -3

p(-3) = (-3)³ − 19(-3) − 30

= -27 + 57 - 30

= -57 + 57

= 0

x² - 3x - 10 = 0

x² - 5x + 2x - 10 = 0

x(x - 5) + 2(x - 5) = 0

(x + 2)(x - 5) = 0

x = -2 and x = 5

So, the solutions are -3, -2 and 5.

Question 8 :

f(x) = 𝑥⁴ − 4𝑥³ - 6𝑥² + 36x − 27 has factor of (x - 3) with multiplicity of two.

Solution :

f(x) = 𝑥⁴ − 4𝑥³ - 6𝑥² + 36x − 27

Since the factor x - 3 has multiplicity of 2, 

x² + 2x - 3 = 0

x² + 3x - 1x - 3 = 0

x(x + 3) -1(x + 3) = 0

(x - 1)(x + 3) = 0

x = 1 and x = -3

So, the solutions are -3, 1, 3 and 3 (since it has the multiplicity of 3).

Question 9 :

When 𝑥³ + 𝑥²- kx - 5 is divided by x - 2 the remainder is 1. Find the value of k.

Solution :

Let p(x) = 𝑥³ + 𝑥²- kx - 5

Dividing by x - 2 the remainder is 1

x - 2 = 0

x = 2

p(2) = 1

p(2) = 2³ + 2²- k(2) - 5

1 = 8 + 4 - 2k - 5

1 = 12 - 5 - 2k

1 = 7 - 2k

-2k = 7 - 1

-2k = 6

k = -3

So, the value of k is -3.

Question 10 :

When the polynomial m𝑥³ - n𝑥² + 5x - 1 is divided by x + 2 the remainder is -39, when the polynomial by x - 1 the remainder is 3. Find the values of m and n.

Solution :

Let p(x) = m𝑥³ - n𝑥² + 5x - 1

x + 2 = 0

x = -2

p(-2) = -39

m(-2)³ - n(-2)² + 5(-2) - 1 = -39

-8m - 4n - 10 - 1 = -39

-8m - 4n - 11 = -39

-8m - 4n = -39 + 11

-8m - 4n = -28

Dividing by -4, we get

2m + n = 7 ------(1)

x - 1 = 0

x = 1

p(1) = 3

m(1)³ - n(1)² + 5(1) - 1 = 3

m - n + 5 - 1 = 3

m - n + 4 = 3

m - n = 3 - 4

m - n = -1 ------(2)

(1) + (2)

2m + n + m - n = 7 - 1

3m = 6

m = 2

Applying the value of m, we get

2 - n = -1

n = 2 + 1

n = 3

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