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CBSE - Class 10 Mathematics Polynomials Worksheet
If one of the zeroes of the quadratic polynomial (k – 1)x2 + kx + 1 is –3, then the value of k is
4/3
b.-4/3
c.2/3
d.-2/3
Choose the correct option: The coefficient of x in the expansion of (x+3)3 is ________.
What is an example of a 4th degree polynomial with exactly 4 terms?
If α and β are the zeroes of the polynomial x2−5x+6, then α + β is:
6
b.5
c.-5
d.-6
If (x + 1) is a factor of x2− 3ax +3a − 7, then the value of a is:
1
b.-1
c.0
d.-2
Given that two of the zeroes of the cubic poly-nomial ax3 + bx² + cx + d are 0, the third zero is
-b/a
b.b/a
c.c/a
d.-d/a
Q2. At what time will she touch the water in the pool?
30 seconds
b.2 seconds
c.1.5 seconds
d.0.5 seconds
If 1 is one of the zeroes of the polynomial x² + x + k, then the value of k is
2
b.-2
c.4
d.-4
How many real roots can a fourth degree polynomial have?
The value of the polynomial x3−3x2+2x+1 at x = -1 is:
0
b.1
c.-3
d.-5
The sum of zeroes of 3x²−8x+4 is:
–8/3
b.4/3
c.–4/3
d.8/3
Choose the correct option: x = 1 is the factor of polynomial _______.
Which one of the following is a polynomial?
The number of polynomials having zeroes as 4 and 7 is
1
b.2
c.4
d.More than 4
If sum of the squares of zeroes of the quadratic polynomial 6x2 + x + k is 25/36, the value of k is:
4
b.-4
c.2
d.-2
What is the number(s) of zeores that a cubic polynomial has/have
0
b.1
c.2
d.3
If the polynomial f(x)=ax3 + bx - c is divisible by g(x)=x2 + bx + c then value of c is equal to?
b
b.2b
c.-2b
If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
10
b.-10
c.5
d.-5
Find the zeros of the polynomial
2, 5
b.-2, 5
c.-2, -5
d.2, -5
Worksheet Answers
An example is -x4 - x3 + 3x + 2.
Solution:
Given: A quadratic polynomial $p(t) = t^2 - 15$.
To Find: The zeroes of the polynomial and verify the relationship between the zeroes and the coefficients of the polynomial.
Step 1: Finding the zeroes of the polynomial
To find the zeroes of the polynomial $p(t)$, we set $p(t) = 0$.
$t^2 - 15 = 0$
We can rewrite this equation using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$.
Since $15 = (\sqrt{15})^2$, we have:
$t^2 - (\sqrt{15})^2 = 0$
$(t - \sqrt{15})(t + \sqrt{15}) = 0$
[By the Zero Product Property, if $ab = 0$, then $a=0$ or $b=0$]
Therefore, $t - \sqrt{15} = 0$ or $t + \sqrt{15} = 0$.
$t = \sqrt{15}$ or $t = -\sqrt{15}$.
The zeroes of the polynomial are $\alpha = \sqrt{15}$ and $\beta = -\sqrt{15}$.
Step 2: Identifying coefficients
Comparing the given polynomial $t^2 - 15$ with the standard quadratic form $at^2 + bt + c$:
$a = 1$
$b = 0$ (since there is no $t$ term)
$c = -15$
Step 3: Verifying the relationship between zeroes and coefficients
The relationships to verify are:
1. Sum of zeroes ($\alpha + \beta$) = $-\frac{b}{a}$
2. Product of zeroes ($\alpha \cdot \beta$) = $\frac{c}{a}$
Verification of Sum of Zeroes:
Sum of zeroes = $\alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0$
Coefficient ratio = $-\frac{b}{a} = -\frac{0}{1} = 0$
Since $0 = 0$, the relationship is verified.
Verification of Product of Zeroes:
Product of zeroes = $\alpha \cdot \beta = (\sqrt{15}) \cdot (-\sqrt{15}) = -(\sqrt{15})^2 = -15$
Coefficient ratio = $\frac{c}{a} = \frac{-15}{1} = -15$
Since $-15 = -15$, the relationship is verified.
Final Answer: The zeroes of the polynomial $t^2 - 15$ are $\sqrt{15}$ and $-\sqrt{15}$. The relationship between the zeroes and coefficients is verified as the sum of zeroes is $0$ and the product of zeroes is $-15$.
0 to 4, inclusive
