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A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form f(x) = a_{3} x^{3 }+ a_{2} x^{2} + a_{1} x + a_{0}. An equation involving a cubic polynomial is called a cubic equation. A closedform solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.
A polynomial of degree 5 is of the form p(x) =, where a, b, c, d, e, and f are real numbers and a ≠ 0.
Thus, p(x) can have at most 6 terms and at least one term containing.
The coefficient of x^{3} in the polynomial 5 + 2x + 3x^{2} – 7x^{3} is
The quadratic polynomial whose sum of zeroes is 3 and the product of zeroes is –2 is :
Sum of zeros = 3/1
b/a = 3/1 .....................(1)
Product of zeros = 2/1
c/a = 2/1 ...................(2)
From equation (1) and (2)
a = 1
b = 3, b = 3
c = 2
The required quadratic equation is
ax^2+by+c
= x^{2}3x2
A number of zeroes of an n degree polynomial = n.
First, a linear polynomial is in the form of ax + b, a≠0, a,b ∈R
The degree of the polynomial = highest degree of the terms
So here the highest degree is 1.
Hence, Linear polynomial has only one zero.
When the polynomial x^{3} + 3x^{2} + 3x + 1 is divided by x + 1, the remainder is :
The zero of x + 1 is –1
And by remainder theorem, when
p(x) = x^{3} + 3x^{2} + 3x + 1 is divided by x + 1, then remainder is p(–1).
∴ p(–1) = (–1)^{3} + 3 (–1)^{2} + 3(–1) + 1
= –1 + (3 × 1) + (–3) + 1
= –1 + 3 – 3 + 1
= 0
Thus, the required = 0
If the polynomial 2x^{3} – 3x^{2} + 2x – 4 is divided by x – 2, then the remainder is :
The value of k for which x – 1 is a factor of the polynomial 4x^{3}+ 3x^{2} – 4x + k is :
X  1 is a factor of 4x^{3} + 3x^{2} 4x +k
then x=1 is one root of 4x^{3} + 3x^{2} 4x +k
put x= 1
4x^{3} +3x^{2} 4x +k = 0
=> 4 (1)^{3} +3 (1)^{2}4 (1) +k =0
=> 4 + 3  4 + k = 0
=> k = 3
The value of k for which x + 1 is a factor of the polynomial x^{3} + x^{2} + x + k is :
The value of m for which x – 2 is a factor of the polynomial x^{4} – x^{3} + 2x^{2} – mx + 4 is :
2x^{2} – 3x – 2
2x^{2}  4x + x  2 = 0
2x^{2}(x2) +1(x2) = 0
(2x+1) (x2)
If x + 2 is a factor of x^{3} – 2ax^{2} + 16, then value of a is
We know that x^{3} + y^{3} + z^{3} – 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx).
If x + y + z = 0, then x^{3} + y^{3} + z^{3} – 3xyz = 0 or x^{3} + y^{3} + z^{3} = 3xyz.
Let x = (a – b), y = (b – c) and z = (c – a)
Consider, x + y + z = (a – b) + (b – c) + (c – a) = 0
⇒ x^{3} + y^{3} + z^{3} = 3xyz
That is (a – b)^{3} + (b – c)^{3} + (c – a)^{3} = 3(a – b)(b – c)(c – a)
The highest power of the variable is known as the degree of the polynomial.
√2x^0 = √2
hence the degree of the polynomial is zero.
The degree of the polynomial 4x^{4}+0x^{3}+0x^{5}+5x+74x^{4}+0x^{3}+0x^{5}+5x+7 is
The degree of the polynomial 4x^{4}+0x^{3}+0x^{5}+5x+74x^{4}+0x^{3}+0x^{5}+5x+7 is
The degree of zero polynomial is not defined, because, in zero polynomial, the coefficient of any variable is zero i.e., Ox^{2 }or Ox^{5}, etc. Hence, we cannot exactly determine the degree of the variable.
Let p (x) = 5x – 4x2 + 3 …(i)
On putting x = 1 in Eq. (i), we get
p(1) = 5(1) 4(1)2 + 3=  5  4 + 3 = 6
p(x)=x+3
p(x)=x+3
p(x)+p(x)=x+3x+3=6
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