Maharashtra Board Class 9 Maths Chapter 3 Set 3.4 Algebra Standard Part 1 Polynomials Solutions

Question 1. For \( x = 0 \), find the value of the polynomial \( x^2 - 5x + 5 \).
Answer: Let \( p(x) = x^2 - 5x + 5 \).
Substituting \( x = 0 \) in the given polynomial:
\( p(0) = (0)^2 - 5(0) + 5 \)
\( \implies p(0) = 0 - 0 + 5 \)
\( \implies p(0) = 5 \)
Therefore, the value of the polynomial for \( x = 0 \) is 5. This simple substitution technique helps us easily find the value of any algebraic expression at a specific point.
In simple words: To find the value, we replace the letter \( x \) with the number 0 everywhere in the expression and then do the basic addition and subtraction to get 5.

🎯 Exam Tip: When substituting 0, remember that any term containing \( x \) becomes 0, leaving only the constant term as the final answer.

Question 2. If \( p(y) = y^2 - 3\sqrt{2}y + 1 \), then find \( p(3\sqrt{2}) \).
Answer: Given polynomial is \( p(y) = y^2 - 3\sqrt{2}y + 1 \).
To find the value, we substitute the given value of the variable into the expression.
Put \( y = 3\sqrt{2} \) in the given polynomial.
\( \therefore p(3\sqrt{2}) = (3\sqrt{2})^2 - 3\sqrt{2}(3\sqrt{2}) + 1 \)
\( = 9 \times 2 - 9 \times 2 + 1 \)
\( = 18 - 18 + 1 \)
\( \therefore p(3\sqrt{2}) = 1 \)
In simple words: To find \( p(3\sqrt{2}) \), we just replace every \( y \) in the equation with \( 3\sqrt{2} \). After multiplying and subtracting the terms, we get the final answer as 1.

🎯 Exam Tip: When squaring a term like \( 3\sqrt{2} \), remember to square both the coefficient and the square root term to get \( 9 \times 2 = 18 \).

Question 3. If \( p(m) = m^3 + 2m^2 - m + 10 \), then \( p(a) + p(-a) = ? \)
Answer: Given polynomial is \( p(m) = m^3 + 2m^2 - m + 10 \).
We will find the expressions for \( p(a) \) and \( p(-a) \) separately before adding them together.
Put \( m = a \) in the given polynomial.
\( \therefore p(a) = a^3 + 2a^2 - a + 10 \) ... (i)
Put \( m = -a \) in the given polynomial.
\( p(-a) = (-a)^3 + 2(-a)^2 - (-a) + 10 \)
\( \therefore p(-a) = -a^3 + 2a^2 + a + 10 \) ... (ii)
Adding equations (i) and (ii),
\( p(a) + p(-a) = (a^3 + 2a^2 - a + 10) + (-a^3 + 2a^2 + a + 10) \)
\( = a^3 - a^3 + 2a^2 + 2a^2 - a + a + 10 + 10 \)
\( \therefore p(a) + p(-a) = 4a^2 + 20 \)
In simple words: We find the value of the polynomial first by putting \( a \) and then by putting \( -a \). When we add both results, the terms with odd powers cancel each other out, leaving us with \( 4a^2 + 20 \).

🎯 Exam Tip: Be very careful with negative signs when raising to a power; a negative number raised to an odd power remains negative, while an even power makes it positive.

Question 4. If \( p(y) = 2y^3 - 6y^2 - 5y + 7 \), then find \( p(2) \).
Answer: Given polynomial is \( p(y) = 2y^3 - 6y^2 - 5y + 7 \).
Substituting a specific number helps us find the numerical value of the polynomial at that point.
Put \( y = 2 \) in the given polynomial.
\( \therefore p(2) = 2(2)^3 - 6(2)^2 - 5(2) + 7 \)
\( = 2 \times 8 - 6 \times 4 - 10 + 7 \)
\( = 16 - 24 - 10 + 7 \)
\( \therefore p(2) = -11 \)
In simple words: We replace every \( y \) in the expression with 2. By following the order of operations (powers first, then multiplication, then addition/subtraction), we calculate the final value to be -11.

🎯 Exam Tip: Always perform exponent operations before multiplying by the coefficient to avoid calculation errors, such as doing \( 2(2)^3 \) as \( 2 \times 8 = 16 \) rather than \( 4^3 \).