Exercise: 12-E
Multiple Choice Questions
Q1: The coefficient of \(y^2\) in \(\frac{-2}{5}xy^2z^3\) is:
Step 1: Identify the given algebraic term.
Given expression is:
\(\frac{-2}{5}xy^2z^3\)
Step 2: Understand the structure of the term.
This is a product of constants and variables: \(\frac{-2}{5} \cdot x \cdot y^2 \cdot z^3\)
Step 3: Definition of coefficient of \(y^2\)
The coefficient of \(y^2\) means: the part of the term that is multiplied with \(y^2\).
So we remove \(y^2\) and keep everything else.
Step 4: Remove \(y^2\) from the term.
Remove \(y^2\) from \(\frac{-2}{5}xy^2z^3\), we are left with: \(\frac{-2}{5}xz^3\)
Answer: d. \(\frac{-2}{5}xz^3\)
Q2: Which of the following expressions is / are polynomials?
I. \(x+\frac{2}{x}\)
II. \(x^3 – y^3 + xy^2 – x^3y\)
III. \(\sqrt{2}x^2 + \sqrt{3}y^2\)
IV. \(7x + 5\sqrt{y} + 3z^2\)
V. \(\sqrt{x^4} + \sqrt{y^2} + z\)
Check Each Expression:
Expression I: \(x + \frac{2}{x}\)
⇒ The term \(\frac{2}{x} = 2x^{-1}\) has a negative exponent.
⇒ So, it is not a polynomial.
Expression II: \(x^3 – y^3 + xy^2 – x^3y\)
⇒ All terms are made up of variables with whole number powers.
⇒ It satisfies all conditions of a polynomial.
⇒ So, it is a polynomial.
Expression III: \(\sqrt{2}x^2 + \sqrt{3}y^2\)
⇒ Square roots of constants like \(\sqrt{2}, \sqrt{3}\) are allowed as coefficients.
⇒ All variables have non-negative integers as exponents.
⇒ So, it is a polynomial.
Expression IV: \(7x + 5\sqrt{y} + 3z^2\)
⇒ The term \(\sqrt{y} = y^{1/2}\) has a fractional exponent.
⇒ Hence, it is not a polynomial.
Expression V: \(\sqrt{x^4} + \sqrt{y^2} + z\)
⇒ \(\sqrt{x^4} = x^2\), \(\sqrt{y^2} = y\)
⇒ All variables now have integer, non-negative exponents.
⇒ So, it is a polynomial.
Answer: b. II, III and V
Q3: Which the following is a binomial?
Step 1: Let’s understand the definition of a binomial.
A binomial is an algebraic expression that contains exactly two unlike terms.
Step 2: Analyze each option.
Option a: \(2x \times 3y = 6xy\)
This is a single term, so it’s a monomial.
Option b: \(3x^2y + 4xy^2\)
Two unlike terms → a valid binomial.
Option c: \(8x^2y + 16yx^2\)
Rewrite both terms: \(8x^2y + 16x^2y = (8 + 16)x^2y = 24x^2y\)
They are like terms, so they simplify to a monomial.
Option d: \(-5x^2yz^2 + 5z^2yx^2\)
Both terms: \(-5x^2yz^2\) and \(5x^2yz^2\)
They are also like terms, so the expression simplifies to: \((-5 + 5)x^2yz^2 = 0\)
Which means the expression becomes 0, not a binomial.
Answer: b. \(3x^2y + 4xy^2\) is a binomial.
Q4: The degree of the polynomial \(a^6 + b^6 – a^3b^4 + a^4b^3\) is
Step 1: Understand what degree means.
The degree of a polynomial is the highest sum of the exponents of variables in any term.
Step 2: Identify the degree of each term.
→ First term: \(a^6\) → Degree = 6
→ Second term: \(b^6\) → Degree = 6
→ Third term: \(-a^3b^4\) → Degree = \(3 + 4 = 7\)
→ Fourth term: \(a^4b^3\) → Degree = \(4 + 3 = 7\)
Step 3: Now find the maximum among all term degrees.
Degrees of terms: 6, 6, 7, 7
Maximum = 7
Answer: d. 7
Q5: The sum of \(\left(6a+4b-c+3\right)\), \(\left(2c-5a-6\right)\), \(\left(2b-3c+4\right)\) and \(\left(11b-7a+2c-1\right)\) is
Step 1: Write all expressions:
\((6a + 4b – c + 3) + (2c – 5a – 6) + (2b – 3c + 4) + (11b – 7a + 2c – 1)\)
Step 2: Group like terms:
= \((6a – 5a – 7a)\) + \((4b + 2b + 11b)\) + \((-c + 2c – 3c + 2c)\) + \((3 – 6 + 4 – 1)\)
Step 3: Simplify each group:
= \(-6a + 17b – 0c + 0\)
Step 4: Final expression:
= \(-6a + 17b\)
Answer: b. -6a + 17b
Q6: \(\left(3q+7p^2-2r^3+4\right)-\left(4p^2-2q+7r^3-3\right)=?\)
Step 1: Write the given expression.
\[
(3q + 7p^2 – 2r^3 + 4) – (4p^2 – 2q + 7r^3 – 3)
\]Step 2: Remove the brackets. Be careful with the minus sign before the second bracket.
\[
= 3q + 7p^2 – 2r^3 + 4 – 4p^2 + 2q – 7r^3 + 3
\]Step 3: Combine like terms.
Group \(p^2\) terms: \(7p^2 – 4p^2 = 3p^2\)
Group \(q\) terms: \(3q + 2q = 5q\)
Group \(r^3\) terms: \(-2r^3 – 7r^3 = -9r^3\)
Group constants: \(4 + 3 = 7\)
Step 4: Final simplified expression:
\[
3p^2 + 5q – 9r^3 + 7
\]Answer: a. \(\left(3p^2 + 5q – 9r^3 + 7\right)\)
Q7: \((x+5)(x-3) = ?\)
Step 1:
Use the identity:
\[
(a + b)(a – b) = a^2 – b^2 \quad \text{(only if both terms are same except sign)}
\]
But here the terms are:
\[
(x + 5)(x – 3) \quad \text{which is not of the form } (a + b)(a – b)
\]
So we expand using distributive property (FOIL method):
Step 2:
\[
(x + 5)(x – 3) = x(x – 3) + 5(x – 3)
\]Step 3:
Multiply each term:
\[
x(x – 3) = x^2 – 3x \\
5(x – 3) = 5x – 15
\]Step 4:
Now add the two results:
\[
x^2 – 3x + 5x – 15
\]Step 5:
Combine like terms:
\[
x^2 + 2x – 15
\]Answer: b. \(x^2 + 2x – 15\)
Q8: (2x + 3)(3x – 1) = ?
Step 1: Multiply the first terms:
(2x) × (3x) = 6x²
Step 2: Multiply the outer terms:
(2x) × (-1) = -2x
Step 3: Multiply the inner terms:
(3) × (3x) = 9x
Step 4: Multiply the last terms:
(3) × (-1) = -3
Step 5: Add all the terms together:
6x² – 2x + 9x – 3
Step 6: Combine like terms:
6x² + 7x – 3
Answer: c. (6x² + 7x – 3)
Q9: \(\left(2x+5\right)\left(2x-5\right)=?\)
Step 1: Observe the expression:
\[
(2x + 5)(2x – 5)
\]Step 2: This is in the form of the algebraic identity:
\[
(a + b)(a – b) = a^2 – b^2
\]Step 3: Compare with identity:
Here, \(a = 2x\) and \(b = 5\)
Step 4: Apply the identity:
\[
(2x)^2 – (5)^2 = 4x^2 – 25
\]Answer: c. \((4x^2 – 25)\)
Q10: \(8x^2y^2\div\left(-2xy\right)=?\)
Step 1: Write the expression
⇒ \( \frac{8x^2y^2}{-2xy} \)
Step 2: Divide constants:
⇒ \( \frac{8}{-2} = -4 \)
Step 3: Apply the law of indices:
\( \frac{x^2}{x} = x^{2-1} = x^1 \)
\( \frac{y^2}{y} = y^{2-1} = y^1 \)
Step 4: Final expression:
⇒ \(-4xy\)
Answer: b. \(-4xy^1\)
Q11: \(\left(2x^2+3x+1\right)\div\left(x+1\right)=?\)
Step 1: We need to divide:
\[
\frac{2x^2 + 3x + 1}{x + 1}
\]We will perform polynomial division.
Step 2: Divide the first term:
\[
\frac{2x^2}{x} = 2x
\]Now multiply \(2x\) with the divisor:
\[
2x(x + 1) = 2x^2 + 2x
\]Subtract:
\[
(2x^2 + 3x + 1) – (2x^2 + 2x) = (3x – 2x) + 1 = x + 1
\]Step 3: Divide the new first term:
\[
\frac{x}{x} = 1
\]Now multiply \(1\) with divisor:
\[
1(x + 1) = x + 1
\]Subtract:
\[
(x + 1) – (x + 1) = 0
\]Step 4: Remainder is 0. Final quotient is:
\[
2x + 1
\]Answer: c. \(\left(2x+1\right)\)
Q12: \(\left(x^2 – 4x + 4\right) \div \left(x – 2\right) = ?\)
Step 1: Factor the numerator:
We recognize that:
\[
x^2 – 4x + 4 = (x – 2)^2
\]Step 2: Rewrite the expression:
\[
\frac{(x – 2)^2}{x – 2}
\]Step 3: Cancel the common factor:
\[
= x – 2
\]Answer: d. \((x – 2)\)
