What must be subtracted from the polynomial 8x^{4} + 14x^{3} + x^{2} + 7x + (8 so that the resulting polynomial is exactly divisible by 4x^{2}  3x + 2?
Thus, when 6x + 2 is subtracted from the given polynomial 8x^{4} + 14x^{3} + x^{2} + 7x + 8, then it will be divisible by 4x^{2}  3x + 2.
Find the smallest solution in positive integers of x^{2}  14y^{2} = 1
Now checking by putting y = 1, 2, 3... until we get solution
y = 1; => x^{2} =15 not possible
y = 2; => x^{2} = 57 not possible
y = 3; => x^{2} = 127 not possible
y = 4; => x^{2} = 225 => x = 15
Hence, the smallest solution is (x, y) = (15, 4).
Obtain all other zeroes of 3x^{4} + 6x^{3}  2x^{2}  10x  5, if two of its zeros are and .
and , x = , x =
= x^{2}  5 / 3 or 3x^{2} – 5 is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial and 3x^{2} – 5.
So, 3x^{4} + 6x^{3} – 2x^{2} – 10x – 5 = (3x^{2} – 5) (x^{2} + 2x + 1) + 0
Quotient = x^{2} + 2x + 1 = (x + 1)^{2}
Zeroes of (x + 1)^{2} are –1, –1.
Hence, all its zeroes are
, , 1, 1
The zeroes of the quadratic polynomial 4x^{2} + 87x + 125 are
using the Quadratic Formula where
a = 4, b = 87, and c = 125
x= −b ± b^{2} − 4ac−−−−−−−√2a
x=−87 ± 872 − 4(4)(125)−−−−−−−−−−−−√2(4)
x= −87 ± 7569 − 2000−−−−−−−−−−√8
x= −87 ± 5569−−−−√8
The discriminant b^{2} − 4ac > 0
so, there are two real roots.
Cannot Simplify the Radical:
x = −87 ± 5569−−−−√8
x = −878 ± 5569−−−−√8
Simplify fractions and/or signs:
x = −878 ± 5569−−−−√8
which becomes x = −1.54678
x = −20.2032
Zeroes of the quadratic polynomial x^{2} + 3x (a + 2)(a  1) is
Zeros of the polynomial obtained by taking P(x) = 0
x^{2} + 3x  (a + 2)(a  1) = 0
x^{2} + (a + 2)x  (a  1)x  (a + 2)(a  1) = 0
x[x + a + 2]  (a  1)[x + a + 2] = 0
[x + a + 2][x  (a  1)] = 0
x = (a + 2) or x = a  1
Hence, the zeros of the polynomial are (a + 2) and (a  1).
The degree of ___ polynomials are not defined.
The value zero" 0" can be considered as a (constant) polynomial, called the zero polynomial.
Like any other constant value, It has no nonzero terms.
There we can say that it has no degree either.
However, its degree is undefined.
Thus a zero is a polynomial of degree zero.
A linear polynomial has one and only one ___.
First, a linear polynomial is in the form of ax + b, a ≠ 0, a, b ∈ R
Degree of the polynomial = highest degree of the terms
So, here the highest degree is 1.
Hence, Linear polynomial has only one zero.
A polynomial of degree n has at most ____ zeroes.
For example: A quadratic equation ax^{2} + bx + c = 0 can have 2 zeros, as the highest power of x is 2 or as the degree is 2. ax^{3} + bx^{2} + cx + d = 0, a cubic equation can have 3 zeros, as the highest power of x is 3 or as the degree is 3.
A polynomial having ____the term is called binomial.
The value of the polynomial f(x) = 4x^{2}  7 at x = 2 is
Given that x = 2
Substitute x in 4x^{2} 7
4(2^{2})  7
4(4)  7
16  7 = 9
If one zero of 3x  8x + 2k + 1 is Severn times the other, then k is____.
a = 3, b = 8 ,c = 2k + 1
suppose the zeroes of the equation are α and β. By the given condition α = 7β
As we know , α + β = b / a
7β + β = (8) / 3
8β = 8 / 3
b = 8 / 3 × 1 / 8 = 1 / 3
αβ = c /a
7β × β = 2k + 1 / 3
7 × 1 / 3 × 1 / 3 = 2k + 1 / 3
7 / 9 = 2k + 1 / 3
21 / 9 = 2k + 1
21 / 9  1 = 2k
12 / 9 = 2k
4 / 3 = 2k
4 / 6 = k
2 / 3 = k
The graph of a quadratic polynomial p(x) = 5x^{2} + 3x + 2 is an upward parabola.
The graph of a quadratic polynomial q(x) =  3x^{2} + 2 is an upward parabola.
If two zeroes of the polynomial (x^{3} − 5x^{2} −16x + 80) are equal in magnitude but opposite in sign, then zeroes are
f(x) = 0 ⇒ x^{3} − 5x^{2} − 16x + 80 = 0
⇒ x^{2} (x − 5) − 16(x − 5) = 0
⇒ (x − 5)(x^{2} − 16) = 0
⇒ (x − 5)(x^{2} − 42) = 0
⇒ (x − 5)(x + 4)(x − 4) = 0
⇒ x = 4, −4, 5
∴ The zeroes of the polynomial are 4, −4, 5
If p(x) is divided by g(x), then deg. p(x) = deg. g(x)  deg. q(x), where q(x) is the quotient.
Here, p(x) = 6x^{2} + 2x + 2g(x) = 2q(x) = 3x^{2} + x + 1 r(x) = 0
The Degrees of p(x) and q(x) are the same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x^{2} + 2x + 2 = 2x (3x^{2} + x + 1)
Hence, the division algorithm is satisfied.
A cubic polynomial must have a minimum of one and maximum of three zeroes.
If two of the zeroes of a cubic polynomial are zero, then it does not contain linear and constant terms.
Linear term of polynomial implies the coefficient of x, (c in the above equation), and the constant term implies the term independent of x, (d in above equation).
Given, two zeroes of the cubic polynomial are zero.
Let two zeros ie.,β,γ = 0. f(x) = (x − α)(x − β)(x − γ)
f(x) = (x − α)(x − 0)(x − 0)
f(x) = (x − α)(x^{2})
f(x) = (x^{3} −x2α)
then the equation does not have a linear term (coefficient of x is 0) and constant term.
If the zeroes of a quadratic polynomial ax^{2} + bx + c are both positive, then a, b and c all have the same sign.
Let α, β are the roots of ax^{2} + bx + c
∴ α + β = b / a and αβ = c / a
Given α and β both are positive.
⇒ b / a is negative.
⇒ a and b both have different signs.
So, the given statement is false.
x^{3}  1 can be the quotient in the division of x^{6} + 2x^{5} + x^{3}  1 by a polynomial of degree 5.
Let p(x) = ax^{2} + bx + c be a quadratic polynomial can have at most
The exponent of the first term is 2.
The exponent of the second term is 1 because bx = bx^{1}.
The exponent of the third term is 0 because c = cx^{0}.
Since the highest exponent is 2, therefore, the degree of ax^{2} + bx + c is 2.
Since, the degree of the polynomial is 2, hence, the polynomial ax^{2} + bx + c can have zero, one or two zeroes.
Hence, the polynomial ax^{2} + bx + c can have at most two zeros.
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