Test: Polynomials - 2 - Class 9 MCQ

The degree of a polynomial is the highest power that is present in it. The degree of the constant polynomial is Zero.

Let a = 3x and b = 5 (3x)³ - (5)³ - 3×3x×5(3x- 5) = 27x³ - 125 -45x ( 3x -5) = 27x³ -125 - 135x² + 225x = 27x³ - 135x² + 225x - 125

X+ 1/x = 7 then,
cubing both side
= (x +1/x)³ = (7)³
= x³ + 1/x³ + 3×x×1/x×(x+1/x) = 343
= x³ +1/x³ +3(7) = 343
= x³ +1/x³ +21 =343
= x³ +1/x³ =343 - 21
= x³ +1/x³ = 322

Degree is the highest exponent of any value in an equation. Here, this highest exponent is 4. Therefore, 4 is the degree of the given equation.

use factor theorem as x+2 is factor of
x³-2ax²+16 so put x = -2 and equate the equation to 0
so putting x = -2
(-2)³-2a(-2)²+16 =0
-8-8a+16=0
-8a = -8
a = 8/8= 1
So,
a = 1

(0.75 * 0.75 * 0.75 + 0.25 * 0.25 * 0.25)/(0.75 * 0.75 - 0.75 * 0.25 + 0.25 * 0.25)

= (0.421875 + 0.015625)/(0.5625 - 0.1875 + 0.0625)

= (0.4375)/(0.4375)

= 1

Let f(x) = px² + 5 x + r
If (x - 2) is a factor of f (x), then by factor theorem
f(2) = 0 | x - 2 = 0 ⇒ x = 2
⇒ p(2)² + 5(2) + r = 0
⇒ 4p + r + 10 = 0    ...(1)

If  is a factor of f (x), then by factor theorem,

Subtracting (2) from (1), we get
3p - 3r = 0
⇒    p = r

It is given that

p(x) = 5x - 4x² + 3

We have to find the value when x = -1

p(-1) = 5(-1) - 4(-1)² + 3

By further calculation

p(-1) = -5 - 4 + 3

So we get

p(-1) = -9 + 3

p(-1) = -6

Therefore, the value is -6.

b-ax=0

b=ax

b/a=x

i.e. remainder is p(b/a)

x³+y³+15xy-125
=x³ + y³ +3 xy ×5 - 125
=x³ + y³ +3xy(x+y) - (5)³
= (x+y) ³ - (5)³
=(5)³ - (5)³
=0