Question 11 Mark
If p(x) = 5x - 4x2 + 3 then p(-1) = ?
- 2
- -2
- 6
- -6
Answer
View full question & answer→- -6
Solution:
P(x) = 5x - 4x2 + 3
⇒ p(-1) = 5(-1) - 4(-1)2 + 3
= -5 - 4 + 3
= -6
32 questions · timed · auto-graded
Solution:
P(x) = 5x - 4x2 + 3
⇒ p(-1) = 5(-1) - 4(-1)2 + 3
= -5 - 4 + 3
= -6
Solution:
Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
p(x) = x2 + x - 6
Now, p(x) = 0
⇒ x2 + x - 6
⇒ x2 + 3x - 2x - 6 = 0
⇒ x(x + 3) - 2(x + 3) = 0
⇒ (x - 2)(x + 3) = 0
⇒ (x - 2) = 0 or (x + 3) = 0
⇒ x = 2 or x = -3
$\therefore$ 2 and -3 are the zeroes of the polynomial p(x).
Solution:
Given, x3 + 2x2 - x - 2
For f(-1),
-1 + 2(-1)2 - (-1) - 2
-1 + 2 -1 - 2 = 0
x = -1,
x + 1 = 0
So, (x + 1) is a factor of the polynomial x3 + 2x2 - x – 2
Solution:
p(x) = x3 + ax2 + 2x +a
x + a = 0 ⇒ x = -a
By the remainder theorem, we know that when p(x) is divided by (x + a), the remainder is p(-a).
Now, p(-a) = x3 + ax2 + 2x +a
= (-a)3 + a(-a)2 + 2(-a) +a
= -a3 + a3 - 2a + a
= -a
Solution:
A polynomial of degree 1 is called a linear polynomial.
Options (a), and (c) have degree 2,
so ther are quadratic polynomials.
option (d) has a negative power, so it is not a polynomial.
The degree of x + 1 is 1, so it is a linear polynomial.
Solution:
Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
p(x) = x2 - 3x
Now, p(x) = 0
⇒ x2 - 3x
⇒ x(x - 3) = 0
⇒ x = 0 or (x - 3) = 0
⇒ x = 0 or x = 3
$\therefore$ 0 and 3 are the zeroes of the polynomial p(x).
Solution:
p(x) = x3 - ax2 + x
x - a = 0 ⇒ x = a
By the remainder theorem, we know that when p(x) is divided by (x - a), the remainder is p(a).
Now, p(a) = a3 - ax2 + a
= a3 - a3 + a
= a
Solution:
p(x) = 2x2 + 7x - 4
Now, p(x) = 0
⇒ 2x2 + 7x - 4 = 0
⇒ 2x2 + 8x - x - 4 = 0
⇒ 2x(x + 4) - 1(x + 4) = 0
⇒ (x + 4)(2x - 1) = 0
⇒ x + 4 = 0 and 2x - 1 = 0
⇒ x = -4 and $\text{x}=\frac{1}{2}$
Solution:
p(x) = 2x3 - kx2 + 3x + 10
x + 2 = 0 ⇒ x = -2
By the factor theorem, we know that when p(x) is divided by (x + 2), the remainder is p(-2).
Now, p(-2) = 2(-2)3 + k(-2)2 + 3(-2) + 10
⇒ 0 = -16 - 4k - 6 + 10
⇒ 0 = -12 - 4k
⇒ 4k = -12
⇒ k = -3
Solution:
Clearly, $\sqrt2\text{x}^2-\sqrt3\text{x}+6$ is a polynomial in one variable because it has only non-negative integral powers of x.
Solution:
The degree of a constant polynomial is 0.
So, $\sqrt3$ is a polynomial of degree 0.
Solution:
Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
A polynomial consisting of one term, namely zero only, is called a zero polynomoial.
So, the zero of a zero polynomial is not defined.
Solution:
Let f(x) = x3 + 10x2 + mx + n
Now, x + 2 = 0 ⇒ x = -2
and x - 1 = 0 ⇒ x = 1
By factor theorem,
f(-2) = 0
⇒ (-2)3 + 10(-2)2 + m(-2) + n
⇒ -8 + 40 - 2m + n = 0
⇒ 2m - n = 32 ...(i)
By factor theorem,
f(1) = 0
⇒ (1)3 + 10(1)2 + m(1) + n = 0
⇒ m + n = -11 ...(ii)
Adding (i) and (ii), we get
3m = 21
⇒ m = 7
Substituting in (ii), we get
n = -18
Solution:
A polynomial consisting of one term, namely zero only, is called a zero polynomial.
So, a zero polynomial can be defined as p(x) = 0.
This can also be written as p(x) = 0 = 0x = 0x2 = 0x3 and so on.
So, it is not possible to determine the degree.
Hence, the degree of a zero polynomial is not defined.
Solution:
Let f(x) = x51 + 51
By the remainder theorem, when f(x) is divided by (x + 1), the remainder is f(-1).
Now, f(-1) = [(-1)n + 51]
= -1 + 51 = 50
Solution:
p(x) = 2x + 5
Now, p(x) = 0
⇒ 2x + 5 = 0
⇒ 2x = -5
$\Rightarrow\text{x}=-\frac{5}{2}$
Solution:
$\text{p}(\text{x})=\text{x}^2-2\sqrt2\text{x}+1$
$\text{p}(2\sqrt2)=(2\sqrt2)^2-2\sqrt2(2\sqrt2)+1$
$=8-8+1$
$=1$
Solution:
p(x) = x3 - 3x2 + 4x + 32
x + 2 = 0 ⇒ x = -2
By the renainder theorem, we know that when p(x) is divided by
(x + 2), the remainder is p(-2).
Now, p(-2) = x3 - 3x2 + 4x + 32
= (-2)3 - 3(-2)2 + 4(-2) + 32
= -8 - 12 - 8 + 32
= 4
Solution:
A polynomial with two non-zero terms is called a binomial.
x2 + 4 is the polynomial that has two non-zero terms.
Hence is a binomial.
Solution:
A polynomial of degree 2 is called a quadratic polynomial.
Options (a), (b) and (c) have degrees 1, 3 and 3 respectively,
so they are not quadratic polynomials.
The degree of x2 + 5x + 4 is 2, so it is a quadratic polynomial.
Solution:
Let p(x) = 2x2 + kx
Since (x + 1) is a factor of p(x),
= P(-1) = 0
⇒ 2(-1)2 + k(-1) = 0
⇒ 2 - k = 0
⇒ k = 2
Solution:
$\text{p}(\text{x}) = 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$\text{x}-1=0\Rightarrow\text{x}=\frac{1}{2}$
By the remainder theorem, we know that when p(x) is divided by (2x - 1), the remainder is $\text{p}\Big(\frac{1}{2}\Big).$
Now, $\text{p}\Big(\frac{1}{2}\Big)= 4\text{x}^3 - 12\text{x}^2 + 11\text{x} - 5$
$=4\Big(\frac{1}{2}\Big)^3-12\Big(\frac{1}{2}\Big)^2+11\Big(\frac{1}{2}\Big)-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$
Solution:
Let p(x) be a polynomial. If $\text{p}(\alpha)=0,$ then we say that $\alpha$ is a zero of a polynomial.
p(x) = 2x2 + 5x -3
Now, p(x) = 0
⇒ 2x2 + 5x -3 = 0
⇒ 2x2 + 6x - x - 3 = 0
⇒ 2x(x + 3) - 1(x + 3) = 0
⇒ (2x - 1)(x + 3) = 0
⇒ (2x - 1) = 0 or (x + 3) = 0
$\Rightarrow\text{x}=\frac{1}{2}$ or x = -3
$$$\therefore\frac{1}{2}$ and -3 are the zeroes of the polynomial p(x).
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (c) have negative and non-integral powers,
So they are not polynomials.
We know that, exery real number is a constant polynomial.
So, -4 being a real number is a polynomial.
Solution:
p(x) = x + 4
p(-x) = -x + 4
p(x) + p(-x) = (x + 4) + (-x + 4)
= x + 4 - x + 4
= 8
Solution:
p(x) = x100 + 2x99 + k
x + 1 = 0 ⇒ x = -1
By the factor theorem, we know that when p(x) is divided by (x + 1), the remainder is p(-1).
Now, p(-1) = (-1)100 + 2(-1)99 + k
⇒ 0 = 1 - 2 + k ...(Given that p(x) is divisible by x + 1.)
⇒ k = 1
Solution:
p(x) = 3x2 - 1
Now, p(x) = 0
⇒ 3x2 - 1 = 0
⇒ 3x2 = 1
$\Rightarrow\text{x}^2=\frac{1}{3}$
$\Rightarrow\text{x}=\frac{1}{\sqrt3}\ \text{and}\ -\frac{1}{\sqrt3}$
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (d) have negative and non-integral powers,
So they are not polynomials.
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (c) have negative and non-integral powers,
So they are not polynomials.
In option (d),
$\text{x}^2+\frac{2\text{x}^\frac{3}{2}}{\sqrt{\text{x}}}+6=\text{x}^2+2\text{x}^{\frac{3}{2}-\frac{1}{2}}+6$
x2 + 2x1 + 6 Which is a polynomial.
Solution:
A polynomial is an algebraic expression in which the variables involved have only non-negative integrals powers.
Option (a), (b) and (c) have negative and non-integral powers,
So they are not polynomials.
We know that, exery real number is a constant polynomial.
So, 0 being a real number is a polynomial.
Solution:
p(x) = x4 + 2x3 - 3x2 + x - 1
x - 2 = 0 ⇒ x = 2
By the remainder theorem, we know that when p(x) is divided by
(x - 2), the remainder is p(2).
Now, p(2) = x4 + 2x3 - 3x2 + x - 1
= (2)4 + 2(2)3 - 3(2)2 + 2 - 1
= 16 +16 - 12 + 2 - 1
= 21
Solution:
p(x) = x3 - 20x + 5k
Now, x + 5 = 0 ⇒ x = (-5)
By factor theorem,
p(-5) = 0
⇒ (-5)3 - 20(-5) + 5k = 0
⇒ -125 + 100 + 5k = 0
⇒ -25 + 5k = 0
⇒ 5k = 25
⇒ k = 5