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Detailed Chapter 6 Factorisation of Algebraic Expressions Set 6.1 MSBSHSE Solutions for Class 8 Maths
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Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions Set 6.1 MSBSHSE Solutions PDF
Question 1. Factorize:
(i) x² + 9x + 18
(ii) x² - 10x + 9
(iii) y² + 24y + 144
(iv) 5y² + 5y - 10
(v) p² - 2p - 35
(vi) p² - 7p - 44
(vii) m² - 23m + 120
(viii) m² - 25m + 100
(ix) 3x² + 14x + 15
(x) 2x² + x - 45
(xi) 20x² - 26x + 8
(xii) 44x² - x - 3
Answer:
(i) x² + 9x + 18
= x² + 6x + 3x + 18
= x (x + 6) + 3(x + 6)
= (x + 6) (x + 3)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 18 दो संख्याओं का गुणनफल है, जबकि नीचे 9 उन्हीं दो संख्याओं का योगफल है। 6 और 3 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: To factorize a quadratic expression like \(x^2 + bx + c\), we find two numbers that multiply to \(c\) and add up to \(b\), then rewrite the middle term and factor by grouping.
🎯 Exam Tip: Always check your factorization by multiplying the binomials back to ensure you get the original expression. This helps verify your answer.
(ii) x² - 10x + 9
= x² - 9x - x + 9
= x (x - 9) - 1(x - 9)
= (x - 9)(x - 1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 9 दो संख्याओं का गुणनफल है, जबकि नीचे -10 उन्हीं दो संख्याओं का योगफल है। -9 और -1 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For \(x^2 - 10x + 9\), we look for two numbers that multiply to 9 and add to -10. These numbers are -9 and -1, which are then used to split the middle term and factor.
🎯 Exam Tip: Pay close attention to the signs of the numbers when finding factors. A common mistake is to get the magnitude right but the sign wrong.
(iii) y² + 24y + 144
= y² + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12)(y + 12)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 144 दो संख्याओं का गुणनफल है, जबकि नीचे 24 उन्हीं दो संख्याओं का योगफल है। 12 और 12 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: This is a perfect square trinomial where the constant term is the square of half the coefficient of the middle term. It factors into \((y+12)^2\).
🎯 Exam Tip: Recognize perfect square trinomials: \(a^2 + 2ab + b^2 = (a+b)^2\) or \(a^2 - 2ab + b^2 = (a-b)^2\). This can save time in factorization.
(iv) 5y² + 5y - 10
= 5(y² + y - 2)
[Taking out the common factor 5]
= 5(y² + 2y - y - 2)
= 5[y(y + 2) - 1(y + 2)]
= 5 (y + 2)(y - 1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर -2 दो संख्याओं का गुणनफल है, जबकि नीचे 1 उन्हीं दो संख्याओं का योगफल है। 2 और -1 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: First, common factors should be taken out. Then, factorize the remaining quadratic expression by finding two numbers that multiply to -2 and add to 1.
🎯 Exam Tip: Always look for a common factor first in any expression before attempting other factorization methods. This simplifies the numbers and makes the process easier.
(v) p² - 2p - 35
= p² - 7p + 5p - 35
= p(p - 7) + 5(p - 7)
= (p - 7)(p + 5)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर -35 दो संख्याओं का गुणनफल है, जबकि नीचे -2 उन्हीं दो संख्याओं का योगफल है। -7 और 5 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For \(p^2 - 2p - 35\), we need to find two numbers that multiply to -35 and add up to -2. These numbers are -7 and 5.
🎯 Exam Tip: When the constant term is negative, the two factors will have opposite signs. The sign of the middle term will tell you which factor (the larger or smaller absolute value) is positive or negative.
(vi) p² - 7p - 44
= p² - 11p + 4p - 44
= p(p - 11) + 4(p - 11)
= (p - 11)(p + 4)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर -44 दो संख्याओं का गुणनफल है, जबकि नीचे -7 उन्हीं दो संख्याओं का योगफल है। -11 और 4 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: To factor \(p^2 - 7p - 44\), find two numbers that multiply to -44 and sum to -7. These are -11 and 4, which helps split the middle term for grouping.
🎯 Exam Tip: Systematically list pairs of factors for the constant term and check their sum until you find the correct pair that matches the middle term's coefficient.
(vii) m² - 23m + 120
= m² - 15m - 8m + 120
= m (m - 15) - 8 (m - 15)
= (m - 15) (m - 8)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 120 दो संख्याओं का गुणनफल है, जबकि नीचे -23 उन्हीं दो संख्याओं का योगफल है। -15 और -8 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For \(m^2 - 23m + 120\), we look for two negative numbers that multiply to 120 and add to -23. These numbers are -15 and -8.
🎯 Exam Tip: If the constant term is positive and the middle term is negative, both factors will be negative. This narrows down the possibilities when searching for factors.
(viii) m² - 25m + 100
= m² - 20m - 5m + 100
= m(m - 20) - 5(m - 20)
= (m - 20) (m - 5)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 100 दो संख्याओं का गुणनफल है, जबकि नीचे -25 उन्हीं दो संख्याओं का योगफल है। -20 और -5 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: To factor \(m^2 - 25m + 100\), find two numbers that multiply to 100 and add to -25. The numbers are -20 and -5.
🎯 Exam Tip: Remember that the order of the terms in the grouping step doesn't affect the final answer, as long as the grouping is done correctly.
(ix) 3x² + 14x + 15
3 × 15 = 45
= 3x² + 9x + 5x + 15
= 3x(x + 3) + 5(x + 3)
= (x + 3) (3x + 5)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 45 (पहले और अंतिम पद का गुणनफल) दो संख्याओं का गुणनफल है, जबकि नीचे 14 उन्हीं दो संख्याओं का योगफल है। 9 और 5 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For trinomials like \(ax^2 + bx + c\) where \(a \neq 1\), multiply \(a\) and \(c\). Find two numbers that multiply to \(ac\) and add to \(b\), then split the middle term and factor by grouping.
🎯 Exam Tip: When the leading coefficient is not 1, use the "splitting the middle term" method by finding factors of \(ac\) that sum to \(b\).
(x) 2x² + x - 45
2 × (- 45) = -90
= 2x² + 10x - 9x - 45
= 2x(x + 5) - 9 (x + 5)
= (x + 5) (2x - 9)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर -90 (पहले और अंतिम पद का गुणनफल) दो संख्याओं का गुणनफल है, जबकि नीचे 1 उन्हीं दो संख्याओं का योगफल है। 10 और -9 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For \(2x^2 + x - 45\), find two numbers that multiply to \(2 \times -45 = -90\) and add to 1. These numbers are 10 and -9.
🎯 Exam Tip: Be careful with negative products and sums. One factor will be positive and one negative when the product \(ac\) is negative.
(xi) 20x² - 26x + 8
= 2(10x² - 13x + 4)
[Taking out the common factor 2]
10 × 4 = 40
= 2(10x² - 8x - 5x + 4)
= 2[2x(5x - 4) - 1(5x - 4)]
= 2 (5x - 4) (2x - 1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर 40 (10 और 4 का गुणनफल) दो संख्याओं का गुणनफल है, जबकि नीचे -13 उन्हीं दो संख्याओं का योगफल है। -8 और -5 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: First, factor out the common factor 2. Then, for the remaining trinomial, find two numbers that multiply to \(10 \times 4 = 40\) and add to -13. These are -8 and -5.
🎯 Exam Tip: Always remember to include the common factor (if any) in your final factored answer. It's a crucial part of the complete factorization.
(xii) 44x² - x - 3
44 × (-3) = -132
= 44x² - 12x + 11x - 3
= 4x(11x - 3) + 1(11x - 3)
= (11x - 3) (4x + 1)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गुणनखंड विधि दर्शाता है। ऊपर -132 (पहले और अंतिम पद का गुणनफल) दो संख्याओं का गुणनफल है, जबकि नीचे -1 उन्हीं दो संख्याओं का योगफल है। -12 और 11 वे संख्याएँ हैं जो इन शर्तों को पूरा करती हैं।
In simple words: For \(44x^2 - x - 3\), multiply \(44 \times -3 = -132\). Find two numbers that multiply to -132 and add to -1. These are -12 and 11.
🎯 Exam Tip: For larger coefficients, systematically checking factors of \(ac\) can be time-consuming. Look for factors that are relatively close to each other if the middle term coefficient is small, or far apart if it's large.
MSBSHSE Solutions Class 8 Maths Chapter 6 Factorisation of Algebraic Expressions Set 6.1
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Detailed Explanations for Chapter 6 Factorisation of Algebraic Expressions Set 6.1
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