UP Board Solutions Class 9 Maths Chapter 5 Polynomial and their Factors Ex 5.4

Balaji Class 9 Maths Solutions Chapter 5 Polynomial And Their Factors Ex 5.4 बहुपद तथा उनके गुणनखण्ड

Exercise 5.4 Polynomial And Their Factors अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Question)

Question 1. 8x + 12y के गुणनखण्ड ज्ञात कीजिए ।
Answer: हलः 8x + 12y = 2 × 2 × 2 ×x+2×2×3xy = 4(2x + 3y)
In simple words: This problem involves finding the greatest common factor (GCF) of the terms and factoring it out from the expression.

🎯 Exam Tip: Always look for the largest common numerical and variable factor in all terms to simplify expressions efficiently.

Question 2. x3 + x2 के गुणनखण्ड ज्ञात कीजिए।
Answer: हलः
x3 + x2 = x x x x x + x x x = x x x(x + 1) = x²(x + 1)
In simple words: To factorize, identify the lowest power of the common variable (x²) and take it out, leaving the remaining terms inside the parenthesis.

🎯 Exam Tip: Remember that x² is a common factor in both x³ and x², simplifying the expression through factorization.

Question 3. x2 + x + y + xy के गुणनखण्ड ज्ञात कीजिए।
Answer: हलः
x2 + x + y + xy = [x2 + x] + [y + xy] = x(x + 1) + y(x + 1) = (x + 1)(x + y)
In simple words: Group terms with common factors, factor them out, and then factor the common binomial.

🎯 Exam Tip: Grouping terms strategically can reveal common binomial factors, which is a key technique for factorization.

Question 4. x2 - x + y - xy के गुणनखण्ड ज्ञात कीजिए ।
Answer: हलः
x2 - x + y - xy = x(x - 1) - y(x - 1) = (x - 1)(x - y)
In simple words: Arrange the terms, factor out common elements from pairs, and then factor the repeating binomial factor.

🎯 Exam Tip: Pay close attention to signs when factoring out negative terms to ensure the binomial factors match.

Exercise 5.4 Polynomial And Their Factors लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

निम्न बहुपदों के गुणनखण्ड ज्ञात कीजिए

Question 5. ax + cx - ay - cy
Answer: हलः ax + cx - ay - cy = x(a + c) - y(a + c) = (a + c)(x - y)
In simple words: Group the terms, factor out the common variable from each pair, and then factor the common binomial (a+c).

🎯 Exam Tip: Identify common factors by careful inspection of each pair of terms before grouping.

Question 6. mx - 2my - nx + 2ny
Answer: हल: mx - 2my - nx + 2ny = mx - nx - 2my + 2ny = x(m - n) - 2y(m - n) = (m - n)(x - 2y)
In simple words: Rearrange terms to group common factors (x and -2y), factor them out, then factor the common binomial (m-n).

🎯 Exam Tip: Sometimes, rearranging terms is necessary to reveal common factors suitable for grouping and factorization.

Question 7. ab + cd + ad + bc
Answer: हलः ab + cd + ad + bc = ab + ad + cd + bc = a(b + d) + c(b + d)= (b + d)(a + c)
In simple words: Rearrange the terms to group common factors and then factor out the common binomial (b+d).

🎯 Exam Tip: Practice rearranging terms in different orders to find the most efficient factorization path.

Question 8. ax - bx + by + cy - cx - ay
Answer: हलः ax - bx + by + cy - cx - ay = ax - bx - cx - ay + by + cy = x(a - b - c) - y(a - b - c) = (a - b - c)(x - y)
In simple words: Rearrange the six terms into two groups, factor out common variables (x and -y) from each group, and then factor the common trinomial (a-b-c).

🎯 Exam Tip: For expressions with many terms, systematic rearrangement and grouping are crucial for successful factorization.

Question 9. mx + ny - nx - my
Answer: हलः mx - ny - nx + my = mx - nx - my + ny = x(m - n) - y(m - n) = (m - n)(x - y)
In simple words: Rearrange the terms to group 'x' and 'y' factors, then factor out the common binomial (m-n).

🎯 Exam Tip: Recognizing that `ny - my` can be written as `-y(m-n)` is a common trick in these types of problems.

Question 10. ax + cy - ac - xy
Answer: हलः ax + cy - ac - xy = ax - ac - xy + y = a(x - c) - y(x - c) = (x - c)(a - y)
In simple words: Rearrange the terms to group those with common factors, factor out `a` and `-y` respectively, and then factor the common binomial `(x-c)`.

🎯 Exam Tip: Be vigilant for potential OCR errors in the source text; however, for exam purposes, the method of grouping and factoring remains consistent.

Question 11. 9a3b2c - 27a2b3c2 - 36a2b2c3
Answer: हलः
9a3b2c - 27a2b3c2 - 36a2b2c3 = 9a2b2c(a - 3bc - 4c²)
In simple words: Find the greatest common factor (GCF) of all numerical and variable parts across all terms and factor it out.

🎯 Exam Tip: The GCF should include the largest common number and the lowest power of each common variable present in all terms.

Question 12. 8a2b(a - b)2 - 4a2(a - b)³
Answer: हलः
8a2b(a - b)² - 4a2 (a - b)³ = 4a2(a - b)²[2b - (a - b)]
= 4a2 (a - b)²[2b - a + b] = 4a2(a - b)²[3b - a]
In simple words: Identify the common factors `4a²` and `(a-b)²`, factor them out, and then simplify the remaining expression inside the brackets.

🎯 Exam Tip: When dealing with binomial factors raised to powers, the lower power of the binomial is part of the GCF.

Question 13. (p + q + r)2 + x[p + q + r) - y(p + q + r)
Answer: हलः :
(p + q + r)2 + x(p + q + r) - y(p + q + r) = (p + q + r)[p + q + r + x - y]
In simple words: Recognize that `(p+q+r)` is a common factor in all terms and factor it out, leaving the remaining terms inside brackets.

🎯 Exam Tip: Treat a complex binomial or trinomial as a single unit if it appears as a common factor in multiple terms.

Question 14. (p + q) + q)3 + (p + q)2 + (p + q)
Answer: हलः
(p + q) q)3 )3 + (p + q)2 q) )2 + (p + q) = (p + q)[(p + q) q)2 + (p + q) + 1]
In simple words: Identify the common factor `(p+q)` from all terms and factor it out, then simplify the expression inside the brackets.

🎯 Exam Tip: The lowest power of the common binomial `(p+q)` is the GCF to be factored out.

Exercise 5.4 Polynomial And Their Factors लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

निम्न बहुपदों के गुणनखण्ड कीजिए।

Question 15. 7(x + y) y)³ +14(x + y) y)² +28(x + y)
Answer: हलः
7(x + y)3 y)3 + 14(x + y)2 y) )2 + 28(x + y) = 7(x + y)[(x + y) y)² + 2(x + y) + 4]
= 7(x + y)[x2 + y2 + 2xy + 2x + 2y + 4]
In simple words: Find the greatest common factor (GCF) `7(x+y)` and factor it out, then expand and simplify the remaining polynomial inside the brackets.

🎯 Exam Tip: After factoring out the GCF, make sure to expand and simplify any remaining terms within the brackets completely.

Question 16. a3bc + ab³c - abc3
Answer: हलः
a3bc + ab³c - abc3 = abc(a2 + b2 - c²)
In simple words: Identify the common factor `abc` from all terms and factor it out to simplify the expression.

🎯 Exam Tip: The GCF includes the lowest power of each variable common to all terms in the polynomial.

Question 17. (x2 + 3x)2 - 5(x2 + 3x) - y(x2 + 3x) + 5y
Answer: हलः
(x2 + 3x)2 - 5(x2 + 3x) - y(x2 + 3x) + 5y = (x2 + 3x)[x2 + 3x - 5] - y[x2 + 3x - 5]
= [x2 + 3x - 5][x2 + 3x - y]
In simple words: Treat `(x²+3x)` as a single unit, then group terms to factor out `(x²+3x)` and `-y` respectively, leading to a common binomial factor.

🎯 Exam Tip: Substituting a complex term with a temporary variable (e.g., let `P = x²+3x`) can simplify the appearance and make factorization clearer.

Question 18. ab(x2 + y2)+ xy(a² + b²)
Answer: हलः
ab(x² + y²) + xy(a² + b²)= abx² + aby2 + xya2 + xyb2 = abx2 + xya² + aby2 + xyb2 = ax(bx + ay ) + by (ay + bx) = (ax + by)(bx + ay)
In simple words: Expand the expression, rearrange the terms, group them based on common factors, and then factor out the common binomial.

🎯 Exam Tip: After expansion, careful rearrangement of terms is often needed to identify pairs with common factors for grouping.

Question 19. x5 + x4 - 2x2 - 2x
Answer: हलः
x5 + x4 - 2x2 - 2x = x{x4 + x3 - 2x - 2] = x[x³(x + 1) - 2(x + 1)]= x[(x3 - 2)(x + 1)]
In simple words: Factor out the common `x`, then group the remaining terms to find `(x+1)` as a common binomial factor.

🎯 Exam Tip: Always look for the greatest common monomial factor first before attempting other factorization methods like grouping.

Question 20. ab(c² + 1) + c(a² + b²)
Answer: हलः
ab(c² + 1) + c(a² + b²) = abc² + ab + ca² + cb2 = abc2 + ca² + ab + cb2 = ac(bc + a) + b(a + bc) = (bc + a)(ac + b)
In simple words: Expand the expression, rearrange terms to group common factors, then factor out the common binomial `(bc+a)`.

🎯 Exam Tip: Rearranging terms to form recognizable groups is crucial for factorization by grouping, especially with expanded polynomials.

Question 21. x² + \( \frac{1}{x^{2}} \) + 2 - 3x - \( \frac{3}{x} \)
Answer: हलः
\[x^{2}+\frac{1}{x^{2}}+2-3 x-\frac{3}{x}=\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)=\left(x+\frac{1}{x}\right)\left[x+\frac{1}{x}-3\right]\]
In simple words: Recognize that \(x^2 + \frac{1}{x^2} + 2\) is a perfect square \( (x + \frac{1}{x})^2 \), then factor out the common term \( (x + \frac{1}{x}) \).

🎯 Exam Tip: Look for perfect square trinomials (like \(a^2 + b^2 + 2ab\)) as they often simplify complex expressions. Let \(P = x + \frac{1}{x}\) to make the factorization easier to visualize.

Balaji Publications Mathematics Class 9 Solutions