UP Board Solutions Class 9 Maths Chapter 5 Polynomial and their Factors Ex 5.5

Balaji Class 9 Maths Solutions Chapter 5 Polynomial And Their Factors Ex 5.5 बहुपद तथा उनके गुणनखण्ड

Ex 5.5 Polynomial And Their Factors अतिलघु उत्तरीय प्रश्न (Very Short Answer Type Question)

Question 1. \(x^2 + 6x + 9\)
Answer: \(x^2 + 6x + 9 = (x)^2 + 2 \times 3 \times x + (3)^2 = (x + 3)^2\)
In simple words: This problem involves factoring a perfect square trinomial. We identify the square roots of the first and last terms and double their product to match the middle term, leading to the form \((a+b)^2\).

🎯 Exam Tip: Recognizing perfect square trinomials \((a^2 + 2ab + b^2)\) and their factorization to \((a+b)^2\) or \((a-b)^2\) is crucial for quick and accurate solutions.

Question 2. \(x^2 - 14x + 49\)
Answer: \(x^2 - 14x + 49 = (x)^2 - 2 \times 7 \times x + (7)^2 = (x - 7)^2\)
In simple words: This is another perfect square trinomial. We identify that \(x^2\) is \((x)^2\) and \(49\) is \((7)^2\), and the middle term \(-14x\) is \(-2 \times x \times 7\), so it factors as \((x-7)^2\).

🎯 Exam Tip: Pay close attention to the sign of the middle term; it dictates whether the factored form is \((a+b)^2\) or \((a-b)^2\).

Question 3. \(9x^2 - 12x + 4\)
Answer: \(9x^2 - 12x + 4 = (3x)^2 - 2 \times 3x \times 2 + (2)^2 = (3x - 2)^2\)
In simple words: Here, the first term \(9x^2\) is \((3x)^2\) and the last term \(4\) is \((2)^2\). The middle term \(-12x\) is \(-2 \times 3x \times 2\), which confirms it's a perfect square trinomial \((a-b)^2\).

🎯 Exam Tip: When the leading coefficient is not 1, factor it into the 'a' term of \((a \pm b)^2\) before applying the formula.

Question 4. \(x^2 - 18x + 81\)
Answer: \(x^2 - 18x + 81 = (x)^2 - 2 \times 9 \times x + (9)^2 = (x - 9)^2\)
In simple words: This expression is a perfect square trinomial because \(x^2\) is \((x)^2\), \(81\) is \((9)^2\), and \(-18x\) is \(-2 \times x \times 9\), so it simplifies to \((x-9)^2\).

🎯 Exam Tip: Always check if the constant term is a perfect square and if the middle term is twice the product of the square roots of the first and last terms.

Question 5. \(x^2 - 4x + 4\)
Answer: \(x^2 - 4x + 4 = (x)^2 - 2 \times 2 \times x + (2)^2 = (x - 2)^2\)
In simple words: We can factor this as a perfect square trinomial. \(x^2\) is \((x)^2\), \(4\) is \((2)^2\), and the middle term \(-4x\) is \(-2 \times x \times 2\), which results in \((x-2)^2\).

🎯 Exam Tip: Practicing factorization of various forms helps in quickly identifying patterns like perfect square trinomials.

Question 6. \(49 - 64x^2\)
Answer: \(49 - 64x^2 = (7)^2 - (8x)^2 = (7 + 8x)(7 - 8x)\)
In simple words: This is a difference of squares problem. \(49\) is \((7)^2\) and \(64x^2\) is \((8x)^2\). Using the formula \(a^2 - b^2 = (a+b)(a-b)\), we factor it into \((7+8x)(7-8x)\).

🎯 Exam Tip: The difference of squares formula, \(a^2 - b^2 = (a+b)(a-b)\), is fundamental and frequently tested. Make sure to identify both 'a' and 'b' correctly.

Question 7. \(16x^2 - 9y^2\)
Answer: \(16x^2 - 9y^2 = (4x)^2 - (3y)^2 = (4x + 3y)(4x - 3y)\)
In simple words: This expression is a difference of two squares. \(16x^2\) is \((4x)^2\) and \(9y^2\) is \((3y)^2\), so it factors into \((4x+3y)(4x-3y)\) using the identity \(a^2 - b^2 = (a+b)(a-b)\).

🎯 Exam Tip: Remember that both terms in the difference of squares must be perfect squares. If they are not, look for a common factor first.

Question 8. \(5x^2 - 80y^2\)
Answer: \(5x^2 - 80y^2 = 5[x^2 - 16y^2] = 5[(x)^2 - (4y)^2] = 5[(x + 4y)(x - 4y)]\)
In simple words: First, we factor out the common factor \(5\). This leaves us with \(x^2 - 16y^2\), which is a difference of squares \((x)^2 - (4y)^2\). We then apply the difference of squares formula to get the final factorization.

🎯 Exam Tip: Always look for a common factor first before attempting to apply other factorization identities. This simplifies the expression and prevents errors.

Ex 5.5 Polynomial And Their Factors लघु उत्तरीय प्रश्न - I (Short Answer Type Questions - I)

Question 9. \((x + y)(x - y)\) का मान ज्ञात कीजिए।
Answer: हलः \((x + y)(x - y)\) वर्गान्तर सूत्र से \( = x^2 - y^2\)
In simple words: This problem asks us to find the value of the product of two binomials, which directly corresponds to the difference of squares formula, resulting in \(x^2 - y^2\).

🎯 Exam Tip: Understanding the formula \((a+b)(a-b) = a^2 - b^2\) is crucial for quickly expanding such products and is often used in reverse for factorization.

Question 10. \((x + y)^3 - x - y\) के गुणनखण्ड ज्ञात कीजिए।
Answer: हलः
\((x + y)^3 - x - y = (x + y)^3 - (x + y)\)
\( = (x + y)[(x + y)^2 - 1]\)
\( = (x + y)[(x + y)^2 - (1)^2] = (x + y)[(x + y + 1)(x + y - 1)]\)
In simple words: We first factor out the common term \((x+y)\). This leaves us with \((x+y)^2 - 1\), which is a difference of squares \(((x+y)^2 - (1)^2)\). Applying the difference of squares formula, we get the final factored form.

🎯 Exam Tip: When factoring complex expressions, identify common terms or patterns like difference of squares or perfect squares within the expression to simplify the process.

Question 11. निम्न व्यंजकों के गुणनखण्ड ज्ञात कीजिए (i) \(9x^2 - 25y^2\)
(ii) \(16 - 81x^2\)
(iii) \(100x^2 - 81y^2\)
(iv) \((a + b)^2 - 9c^2\)
Answer: हलः
(i) \(9x^2 - 25y^2 = (3x)^2 - (5y)^2 = (3x + 5y)(3x - 5y)\)
(ii) \(16 - 81x^2 = (4)^2 - (9x)^2 = (4 + 9x)(4 - 9x)\)
(iii) \(100x^2 - 81y^2 = (10x)^2 - (9y)^2 = (10x + 9y)(10x - 9y)\)
(iv) \((a + b)^2 - 9c^2 = (a + b)^2 - (3c)^2 = (a + b + 3c)(a + b - 3c)\)
In simple words: All these sub-parts involve factoring expressions using the difference of squares identity, \(a^2 - b^2 = (a+b)(a-b)\). We identify 'a' and 'b' for each expression and then apply the formula.

🎯 Exam Tip: Master the difference of squares identity as it's a versatile tool for factoring many quadratic and polynomial expressions.

Ex 5.5 Polynomial And Their Factors लघु उत्तरीय प्रश्न - II (Short Answer Type Questions - II)

Question 12. निम्न बहुपदों के गुणनखण्ड ज्ञात कीजिए। \(9x^4 - 6x^3b + x^2b^2\)
Answer: हलः
\(9x^4 - 6x^3b + x^2b^2 = x^2[9x^2 - 6xb + b^2] = x^2[(3x)^2 - 2 \times 3x \times b + (b)^2] = x^2[(3x - b)^2]\)
In simple words: First, we factor out the common term \(x^2\). The remaining trinomial \(9x^2 - 6xb + b^2\) is a perfect square trinomial, as \(9x^2 = (3x)^2\), \(b^2 = (b)^2\), and \(-6xb = -2 \times 3x \times b\). Thus, it factors as \((3x-b)^2\).

🎯 Exam Tip: Always begin factorization by finding the greatest common factor (GCF). Then, look for other factoring patterns like perfect square trinomials or difference of squares.

Question 13. \(x^2 - x + \frac{1}{4}\)
Answer: हल: \(x^{2}-x+\frac{1}{4}=x^{2}-2 \times \frac{1}{2} \times x+\left(\frac{1}{2}\right)^{2}=\left(x-\frac{1}{2}\right)^{2}\)
In simple words: This expression is a perfect square trinomial. We recognize that \(x^2\) is \((x)^2\), \(\frac{1}{4}\) is \(\left(\frac{1}{2}\right)^2\), and the middle term \(-x\) is \(-2 \times x \times \frac{1}{2}\), allowing us to factor it as \(\left(x-\frac{1}{2}\right)^2\).

🎯 Exam Tip: Even with fractions, the perfect square trinomial formula holds. Verify if the middle term is twice the product of the square roots of the first and last terms.

Question 14. \((a + b + c)^2 + 2(a + b + c)(a - b - c)+ (a - b - c)^2\)
Answer: हलः
\((a + b + c)^2 + 2(a + b + c) (a - b - c) + (a - b - c)^2 = [(a + b + c) + (a - b - c)]^2\)
\( = [a + b + c + a - b - c]^2\)
\( = [2a]^2 = 4a^2\)
In simple words: This expression is in the form \(A^2 + 2AB + B^2\), where \(A = (a+b+c)\) and \(B = (a-b-c)\). We can factor it as \((A+B)^2\). Substituting A and B back, the \(b\) and \(c\) terms cancel out, leaving \((2a)^2\), which simplifies to \(4a^2\).

🎯 Exam Tip: Recognize complex expressions that fit basic algebraic identities by substituting parts of the expression with single variables (e.g., A and B) to simplify the pattern identification.

Question 15. \(6x^4 - 24x^3y^3 + 24y^6x^2\)
Answer: हलः
\(6x^4 - 24x^3y^3 + 24y^6x^2 = 6x^2[x^2 - 4xy^3 + 4y^6]\)
\( = 6x^2[(x)^2 - 2 \times x \times 2y^3 + (2y^3)^2] = 6x^2(x - 2y^3)^2\)
In simple words: First, factor out the greatest common factor, \(6x^2\). The remaining expression \(x^2 - 4xy^3 + 4y^6\) is a perfect square trinomial, as \(x^2 = (x)^2\), \(4y^6 = (2y^3)^2\), and \(-4xy^3 = -2 \times x \times 2y^3\). Thus, it factors as \((x - 2y^3)^2\).

🎯 Exam Tip: For expressions with multiple variables and higher powers, always start by extracting the largest common factor from all terms. Then, apply appropriate factorization formulas.

Question 16. \(121x^2y^2 + 110xyab + 25a^2b^2\)
Answer: हलः
\(121x^2y^2 + 110xyab + 25a^2b^2 = (11xy)^2 + 2 \times 11xy \times 5ab + (5ab)^2 = (11xy + 5ab)^2\)
In simple words: This expression is a perfect square trinomial. We can see that \(121x^2y^2\) is \((11xy)^2\), \(25a^2b^2\) is \((5ab)^2\), and the middle term \(110xyab\) is \(2 \times 11xy \times 5ab\), which means it factors into \((11xy + 5ab)^2\).

🎯 Exam Tip: When dealing with multiple variables, treat each combined term (like \(xy\) or \(ab\)) as a single unit when checking for perfect square trinomial patterns.

Question 17. \(x^{2} z^{2}+\frac{1}{25} y^{2}-\frac{2}{5} x y z\)
Answer: हल: \(x^{2} z^{2}+\frac{1}{25} y^{2}-\frac{2}{5} x y z=(x z)^{2}+\left(\frac{1}{5} y\right)^{2}-2 \times x z \times \frac{1}{5} y=\left(x z-\frac{1}{5} y\right)^{2}\)
In simple words: This expression is a perfect square trinomial. We identify \((xz)^2\) as the first term squared, \(\left(\frac{1}{5}y\right)^2\) as the last term squared, and the middle term \(-2 \times xz \times \frac{1}{5}y\) which matches \(-2 \times (\text{first term}) \times (\text{second term})\), allowing it to be factored as \(\left(xz - \frac{1}{5}y\right)^2\).

🎯 Exam Tip: Be comfortable with recognizing perfect square trinomials even when terms involve multiple variables or fractions. Identify the square roots of the 'a' and 'b' terms and verify the middle term's structure.

Question 18. \(3x^3y - 243xy^3\)
Answer: हलः
\(3x^3y - 243xy^3 = 3xy[x^2 - 81y^2] = 3xy[(x)^2 - (9y)^2] = 3xy[(x + 9y)(x - 9y)]\)
In simple words: First, we factor out the common term \(3xy\). The remaining expression \(x^2 - 81y^2\) is a difference of squares \((x)^2 - (9y)^2\). Applying the difference of squares formula, we get \((x+9y)(x-9y)\) as the factored form of the bracketed part.

🎯 Exam Tip: Always look for the greatest common factor first. After factoring it out, the remaining expression often simplifies to a standard identity like difference of squares or a perfect square trinomial.

Question 19. \(1 - 2xy - (x^2 + y^2)\)
Answer: हलः
\(1 - 2xy - (x^2 + y^2) = 1 - [2xy + x^2 + y^2] = (1)^2 - [(x + y)^2]\) वर्गान्तर सूत्र से \( = (1 + x + y)(1 - x - y)\)
In simple words: We first rearrange the terms to reveal a perfect square trinomial within the parenthesis: \((x^2 + 2xy + y^2)\) which is \((x+y)^2\). Then the expression becomes a difference of squares, \(1^2 - (x+y)^2\), which factors into \((1 + (x+y))(1 - (x+y))\).

🎯 Exam Tip: Be alert to expressions that can be grouped to form standard identities. Rearranging terms and factoring out negative signs can often expose these patterns.

Question 20. \(x^2 + 6x + 9 - 25y^2\)
Answer: हलः
\(x^2 + 6x + 9 - 25y^2 = x^2 + 2 \times 3 \times x + (3)^2 - (5y)^2\)
\( = (x + 3)^2 - (5y)^2 = (x + 3 + 5y)(x + 3 - 5y)\)
In simple words: The first three terms \((x^2 + 6x + 9)\) form a perfect square trinomial, \((x+3)^2\). The entire expression then becomes a difference of squares, \((x+3)^2 - (5y)^2\), which factors into \((x+3+5y)(x+3-5y)\).

🎯 Exam Tip: Look for combinations of factoring techniques. Often, a perfect square trinomial combines with another term to form a difference of squares, which is a common pattern in advanced factorization.

Question 21. \(2xy - (x^2 + y^2 - z^2)\)
Answer: हलः
\(2xy - (x^2 + y^2 - z^2) = 2xy - x^2 - y^2 + z^2 = z^2 - (x^2 + y^2 - 2xy) = z^2 - (x - y)^2 = (z + x - y)(z - x + y)\)
In simple words: We first distribute the negative sign, then rearrange terms to group \((x^2 - 2xy + y^2)\) which is \((x-y)^2\). The expression then becomes \(z^2 - (x-y)^2\), a difference of squares, factoring into \((z + (x-y))(z - (x-y))\).

🎯 Exam Tip: When a negative sign precedes a parenthesis, remember to change the sign of every term inside the parenthesis. This step is crucial for correctly identifying subsequent factorization patterns.

Question 22. \(x^2 - y^2 - 2x + 1\)
Answer: हलः
\(x^2 - y^2 - 2x + 1 = x^2 - 2x + 1 - y^2\)
\( = (x)^2 - 2 \times x \times 1 + (1)^2 - y^2\)
\( = (x - 1)^2 - (y)^2 = (x - 1 + y)(x - 1 - y)\)
In simple words: We rearrange the terms to group \((x^2 - 2x + 1)\) which is a perfect square trinomial, \((x-1)^2\). The entire expression then becomes a difference of squares, \((x-1)^2 - y^2\), which factors into \((x-1+y)(x-1-y)\).

🎯 Exam Tip: Reordering terms is a powerful strategy to reveal hidden patterns. Look for combinations of terms that form known identities, especially perfect square trinomials, to then simplify the problem into a difference of squares.