Test: Division Algorithm for Polynomials - Class 10 MCQ

Step-by-step explanation:

Given: Dividend = 

Divisor =  x² - 4x + 3

Solution :

Since we know that :

Since the remainder is  2x-3 

So, 2x-3 must be subtracted from     
so that the resulting  polynomial is exactly divisible by x² - 4x + 3

Since (X+1) is a factor of x2 - 3ax + 3a - 7 

Therefore , x+1=0

or, x=-1

Putting the value of x in the the p(x) = x2 - 3ax + 3a - 7 

or, (-1)² -3(a)(-1)+3a-7=0

1+3a+3a-7=0

6a=6

or, a=1

Therefore the  value of a=1.

We put the value x=-2 in the polynomial 2x²-3x-2
f(-2)=2(-2)²-3(-2)-2=2*4+6-2=12

The degree of the divisor is 2

Step-by-step explanation:

x⁵÷x³=x²

By the Division algorithm 155 = 9(17) + 2. Where remainder is 155 mod 9.

Divisor*Quotient+Remainder= Divident
g(x)(x-2)+(-2x+4)=x3-3x2+x+2
g(x)(x-2)=x3-3x2+x+2+2x-4
g(x)(x-2)=x3-3x2+3x-2
g(x)=x3-3x2+3x-2/(x-2)
g(x)=x²-x+1

Since the solution is not a complex number, it is a real number. Since rational numbers include all integers, natural numbers and fractions , So the numbers are rational numbers.

Given is the value of a and b. 

given that (x+1) is a factor of p(x) 

therefore, -1 is a zero of given p(x) 

p(x) = 2x3 +ax2 +2bx +1

 substituting the value of -1 in the given p(x), we get

 p(x)=2*(-1)3 +a *(-1)2 +2*b*(-1)+ 1

 = -2 + a -2b + 1 

= -1 + a - 2b

 or,a - 2b = 1 

also given that 2a - 3b = 4 

so we got two equations; 

a - 2b = 7 ...(1)

 2a - 3b = 4 ...(2) 
(1)* (2)* (4)- 2 = 2a - 4b = (3)

(2)* 1 = 2a - 3b = 4 (4) 

(4) -  (3) = [2a - 2a ] + [-3b - (-4b)] = 4 - 2

-3b + 4b = 2

therefore b = 2 

substituting the value of bin (3)

2a - 4b = 2

2a - (4*2) = 2

2a - 8 = 2

2a = 2 + 8

2 = 10

a = 10 / 2

therefore a = 5 

so we get the value of a and b 

that is ; a = 5 and b = 2

We have the polynomial 2x³+x2-5x+2
By Hit and trial method, x=1 is the solution.
Dividing the polynomial by x-1
We get quotient as = 2x²+3x-2

Given that the remainder is (x + a)
⇒ (4k – 25 + 16 – 2k)x + [10 – k(8 – k) ] = x + a  
⇒ (2k – 9)x + [10 – 8k + k² ] = x + a
On comparing both the sides, we get
2k  – 9 = 1
⇒ 2k = 10
∴ k = 5
Also 10 – 8k + k² = a
⇒ 10 – 8(5) + 5² = a
⇒ 10 – 40 + 25 = a
∴ a = – 5

Degree of remainder is one less than the degree of the divisor because if the remainder is of the same degree then it can be divided further , its just like when the divisor is a number the remainder is less than the number because if it is more than or equal to that number it can further be divided. So, the degree is zero.

let the given polynomial be

since 2x+3=0

if (2x+3) is a factor of f(x). therefore by the factor theorem f(-3/2) =0

Dividend = Divisor × Quotient + Remainder.
When we divide f(x) by g(x) then q(x) is quotient and r(x) is remainder.
Therefore, f(x) = g(x) × q(x) + r(x).

The polynomial with degree 4 is a biquadratic equation. And a polynomial has no. of zeros equal to the degree of the polynomial. So total no. of zeros are 4. 2 zeros are already provided to us , so it has 2 more zeros.