MTH-116 Section 32906 Spring 2019: Expressions and Polynomials

This assignment focuses on algebra, covering topics such as evaluating expressions, simplifying polynomials, and solving equations. The solutions demonstrate the application of algebraic principles, including substitution, exponent rules, and combining like terms. Specific problems involve evaluating expressions with given variable values, identifying and classifying polynomials, performing algebraic manipulations, and solving inequalities. The document provides step-by-step solutions, making it a useful resource for students studying algebra. Desklib offers more solved assignments and past papers for students.

CH – 2
Solution 1: Given , we need to find the value of
Now substitute the value of in equation (1) we get,
Hence
Solution 2: Given , we need to find the value of
Now substitute the value of in equation (1) we get,
Hence
Solution 3: Given , we need to find the value of
Now substitute the value of in equation (1) we get,
Hence
Solution 4:
(a): Since and since so
(b): Since and since so
(c): Since and since so
(d): Since and since so
Solution 5: Given . Suppose that the given expression is S, then
Hence,

Solution 6:Given . Suppose that the given expression is S, then
Hence,
Solution 7: Given . Suppose that the given expression is S, then
Hence,
Solution 8: Given . Suppose that the given expression is S, then
Hence,
Solution 9: Given . Suppose that the given expression is S, then
Hence,
Solution 10:
(a):
(b):
(c):
(d):
Solution 11:
Solution 12:
Solution 13:
Solution 14:
Solution 15:
Solution 16:
Solution 17:
Solution 18:
Solution 19:
Solution 20:

Solution 21:
Solution 22:
Note: Standard form of Fraction: When numerator and denominator are co-prime, then
the fraction is said to be in standard form. Co-prime means greatest common divisor of
two numbers is 1.
(a):
(b):
(c):
(d):
Solution 23:
(a): Since , a rational number where
So is a rational number.
Hence 1st option is correct.
(b): Since , a rational number where
So is a rational number.
Hence 4rth option is correct.
(c): Since , a rational number where
So, is a rational number.
Hence 1st option is correct.
(d): Since , a rational number where
Hence 2nd option is correct.
Solution 24:Suppose that the distance between A and B be x ft as shown in the given
Since triangle ABC is a right angle triangle so use Pythagoras theorem

CH – 3
Solution 1:
(a): Given expression is .
The polynomial is a binomial of degree 3.
(b):
The polynomial is a trinomial of degree 1.
Solution 2:
(a): Given expression is .
We know that
So,
Hence,
(b): Given expression is .
We know that
So,
Hence,
Solution 3:
(a): Given expression is
We know that
So,
Hence,
(b): Given expression is
We know that
So,

Hence,
Solution 4:
(a): Given expression is
(b): Given expression is
Solution 5:Given expression is .
Rewrite the given expression and simplify as,
Hence,
Solution 6: Given expression is .
Rewrite the given expression and simplify as,
Hence,
Solution 7: The given algebraic expression is .
Rewrite the given expression and simplify as,
Hence,
Solution 8:
(a): The given algebraic expression is .
Rewrite the given expression and simplify as,
Hence,
(b): The given algebraic expression is .
Rewrite the given expression and simplify as,
Hence,
Solution 9:
(a): The given algebraic expression is .
Use formula we get,

Hence,
(b): The given algebraic expression is .
Use formula we get,
Hence,
Solution 10:
Hence,
Solution 11:
Hence,
Solution 12:
Hence,
Solution 13:
Hence,
Solution 14:
Hence
Solution 15:
(a):
Hence,
(b):

Hence,
Solution 16:
Hence,
Solution 17:
Hence,
Solution 18:Suppose that the number be x then according to the question,
Now let’s solve
Hence, the required number is
Solution 20:
(a): The graph of the solution of the inequality on number line is shown below,
(b): The graph of the solution of the inequality on number line is shown below,
Solution 20:
The graph of the solution of the inequality on number line is shown below,