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A polynomial containing two nonzero terms is called a ________.
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 of the polynomial 4x^{4} + 0x^{3} + 0x^{5} + 5x + 7 is:
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.
A polynomial containing three nonzero terms is called a ________.
If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is
use factor theorem as x+2 is factor of
x³2ax²+16 so put x = 2 and equate the equation to 0
so putting x = 2
(2)³2a(2)²+16 =0
88a+16=0
8a = 8
a = 8/8= 1
So,
a = 1
ANSWER : b
Solution : (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
If one of the factor of x^{2} + x – 20 is (x + 5). Find the other
Let f(x) = px^{2} + 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
If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is
The remainder when the polynomial x^{4}+2x^{3}−3x^{2}+x−1 is divided by (x−2) is
The remainder obtained when the polynomial p(x) is divided by (b – ax) is
ANSWER : b
Solution : bax=0
b=ax
b/a=x
i.e. remainder is p(b/a)
x³+y³+15xy125
=x³ + y³ +3 xy ×5  125
=x³ + y³ +3xy(x+y)  (5)³
= (x+y) ³  (5)³
=(5)³  (5)³
=0
If the polynomial x^{3}−6x^{2}+ax+3 leaves a remainder 7 when divided by (x−1), then the value of ‘a’ is
The value of (a^{2}−b^{2})^{3}+(b^{2}−c^{2})3+(c^{2}−a^{2})^{3} is
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