RBSE Solutions Class 8 Maths Chapter 10 Factorization More Ques

Question 1. Find the two integers a and b such that

Answer: We use the identity: \( (a - b)^2 = (a + b)^2 - 4ab \)
(i) If \( a+b = 8 \) and \( ab = 15 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(8)^2 - 4 \times 15} \)
\( \implies a - b = \sqrt{64 - 60} \)
\( \implies a - b = \sqrt{4} \)
\( \implies a - b = 2 \) We have \( a+b = 8 \) and \( a-b = 2 \). Adding these equations: \( (a+b) + (a-b) = 8 + 2 \)
\( \implies 2a = 10 \)
\( \implies a = 5 \) Substitute \( a = 5 \) into \( a+b = 8 \): \( 5 + b = 8 \)
\( \implies b = 8 - 5 \)
\( \implies b = 3 \) So, \( a = 5 \) and \( b = 3 \).
(ii) If \( a+b = 13 \) and \( ab = 12 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(13)^2 - 4 \times 12} \)
\( \implies a - b = \sqrt{169 - 48} \)
\( \implies a - b = \sqrt{121} \)
\( \implies a - b = 11 \) We have \( a+b = 13 \) and \( a-b = 11 \). Adding these equations: \( (a+b) + (a-b) = 13 + 11 \)
\( \implies 2a = 24 \)
\( \implies a = 12 \) Substitute \( a = 12 \) into \( a+b = 13 \): \( 12 + b = 13 \)
\( \implies b = 13 - 12 \)
\( \implies b = 1 \) So, \( a = 12 \) and \( b = 1 \).
(iii) If \( a+b = -1 \) and \( ab = -20 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(-1)^2 - 4 \times (-20)} \)
\( \implies a - b = \sqrt{1 + 80} \)
\( \implies a - b = \sqrt{81} \)
\( \implies a - b = 9 \) We have \( a+b = -1 \) and \( a-b = 9 \). Adding these equations: \( (a+b) + (a-b) = -1 + 9 \)
\( \implies 2a = 8 \)
\( \implies a = 4 \) Substitute \( a = 4 \) into \( a+b = -1 \): \( 4 + b = -1 \)
\( \implies b = -1 - 4 \)
\( \implies b = -5 \) So, \( a = 4 \) and \( b = -5 \).
(iv) If \( a+b = -5 \) and \( ab = 4 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(-5)^2 - 4 \times 4} \)
\( \implies a - b = \sqrt{25 - 16} \)
\( \implies a - b = \sqrt{9} \)
\( \implies a - b = 3 \) We have \( a+b = -5 \) and \( a-b = 3 \). Adding these equations: \( (a+b) + (a-b) = -5 + 3 \)
\( \implies 2a = -2 \)
\( \implies a = -1 \) Substitute \( a = -1 \) into \( a+b = -5 \): \( -1 + b = -5 \)
\( \implies b = -5 + 1 \)
\( \implies b = -4 \) So, \( a = -1 \) and \( b = -4 \).
(v) If \( a+b = 10 \) and \( ab = 21 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(10)^2 - 4 \times 21} \)
\( \implies a - b = \sqrt{100 - 84} \)
\( \implies a - b = \sqrt{16} \)
\( \implies a - b = 4 \) We have \( a+b = 10 \) and \( a-b = 4 \). Adding these equations: \( (a+b) + (a-b) = 10 + 4 \)
\( \implies 2a = 14 \)
\( \implies a = 7 \) Substitute \( a = 7 \) into \( a+b = 10 \): \( 7 + b = 10 \)
\( \implies b = 10 - 7 \)
\( \implies b = 3 \) So, \( a = 7 \) and \( b = 3 \).
(vi) If \( a+b = -1 \) and \( ab = -12 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(-1)^2 - 4 \times (-12)} \)
\( \implies a - b = \sqrt{1 + 48} \)
\( \implies a - b = \sqrt{49} \)
\( \implies a - b = 7 \) We have \( a+b = -1 \) and \( a-b = 7 \). Adding these equations: \( (a+b) + (a-b) = -1 + 7 \)
\( \implies 2a = 6 \)
\( \implies a = 3 \) Substitute \( a = 3 \) into \( a+b = -1 \): \( 3 + b = -1 \)
\( \implies b = -1 - 3 \)
\( \implies b = -4 \) So, \( a = 3 \) and \( b = -4 \).
(vii) If \( a+b = -11 \) and \( ab = 10 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(-11)^2 - 4 \times 10} \)
\( \implies a - b = \sqrt{121 - 40} \)
\( \implies a - b = \sqrt{81} \)
\( \implies a - b = 9 \) We have \( a+b = -11 \) and \( a-b = 9 \). Adding these equations: \( (a+b) + (a-b) = -11 + 9 \)
\( \implies 2a = -2 \)
\( \implies a = -1 \) Substitute \( a = -1 \) into \( a+b = -11 \): \( -1 + b = -11 \)
\( \implies b = -11 + 1 \)
\( \implies b = -10 \) So, \( a = -1 \) and \( b = -10 \).
(viii) If \( a+b = -7 \) and \( ab = 10 \) Then, \( a - b = \sqrt{(a+b)^2 - 4ab} \)
\( \implies a - b = \sqrt{(-7)^2 - 4 \times 10} \)
\( \implies a - b = \sqrt{49 - 40} \)
\( \implies a - b = \sqrt{9} \)
\( \implies a - b = 3 \) We have \( a+b = -7 \) and \( a-b = 3 \). Adding these equations: \( (a+b) + (a-b) = -7 + 3 \)
\( \implies 2a = -4 \)
\( \implies a = -2 \) Substitute \( a = -2 \) into \( a+b = -7 \): \( -2 + b = -7 \)
\( \implies b = -7 + 2 \)
\( \implies b = -5 \) So, \( a = -2 \) and \( b = -5 \).
Here is the completed table:

In simple words: We are given the sum and product of two numbers, \(a\) and \(b\). We use a special formula that relates \(a+b\), \(ab\), and \(a-b\). Once we find \(a-b\), we can easily solve for \(a\) and \(b\) by adding and subtracting the two resulting equations. This method helps to find the individual numbers.
🎯 Exam Tip: Remember the identity \( (a - b)^2 = (a + b)^2 - 4ab \) for solving problems where the sum and product of two variables are given. It's a key formula in factorization.

Question 2. Find the error.
Solution: \( 3x + x + 4x = 56 \)
\( \implies 3x + 1x + 4x = 56 \)
\( \implies (3 + 1 + 4) x = 56 \)
\( \implies 8x = 56 \)
\( \implies x = \frac {56}{8} = 7 \) (correct value)
Answer: The question asks to "Find the error" but then provides a completely correct step-by-step solution. There is no error in the provided calculation. All the steps correctly simplify the equation \( 3x + x + 4x = 56 \) to find \( x = 7 \). This shows clear and accurate algebraic steps.
In simple words: The math steps given are all correct, so there is no mistake in how \(x\) was found. The problem statement itself implies an error, but the working is perfect.
🎯 Exam Tip: When asked to find an error, first carefully follow each step of the given solution. If no error is found, state that the solution is correct.

Question 3. Find the value of \( 5x \) at \( x = -2 = 5-2 = 3 \). Find the error and also find the correct value.
Answer: The error in the question statement is that \( x = -2 \) cannot be equal to \( 5-2=3 \) at the same time. The values \(-2\) and \(3\) are different. Assuming the intention was to find \( 5x \) when \( x = -2 \):
Value of \( 5x \) at \( x = -2 \)
\( = 5 \times (-2) \)
\( = -10 \) If the intention was to use \( x = 3 \) (from \( 5-2 \)):
Value of \( 5x \) at \( x = 3 \)
\( = 5 \times 3 \)
\( = 15 \)
The question mixes two different values for \(x\). We usually solve for \(x\) based on the most direct statement, which is \(x = -2\).
In simple words: The mistake is saying \(x\) is both \(-2\) and \(3\) at once, which is not possible. If \(x\) is \(-2\), then \(5x\) is \(-10\). If \(x\) is \(3\), then \(5x\) is \(15\).
🎯 Exam Tip: Always be careful when a question gives conflicting information. Identify the clear statements and resolve the ambiguity before calculating, or show results for each possible interpretation.

Question 4. Solutions of the expression is given in the column A and B. Find which of the solution is correct.

Answer: (i) For \( 3(x-4) \) Using the distributive property, \( 3 \times x - 3 \times 4 = 3x - 12 \). Hence, B is correct.
(ii) For \( (2x)^2 \) \( (2x)^2 = (2x) \times (2x) = (2 \times x) \times (2 \times x) = 2 \times 2 \times x \times x = 4x^2 \). Hence, B is correct.
(iii) For \( (x+4)^2 \) Using the identity \( (a+b)^2 = a^2+2ab+b^2 \), we get \( x^2 + 2(x)(4) + (4)^2 = x^2+8x+16 \). Hence, B is correct.
(iv) For \( (x-3)^2 \) Using the identity \( (a-b)^2 = a^2-2ab+b^2 \), we get \( x^2 - 2(x)(3) + (3)^2 = x^2-6x+9 \). Hence, B is correct.
(v) For \( \frac {y+5}{5} \) (assuming the question intended \( \frac{y+5}{5} \) instead of \( \frac{y+1}{5} \) based on solution and visual) \( \frac {y+5}{5} = \frac {y}{5} + \frac {5}{5} = \frac {y}{5} + 1 \). Hence, B is correct.
In simple words: For each math problem, we look at the two possible answers, A and B. We then work out the problem step-by-step using the correct math rules. For all the given expressions, option B gives the correct simplified or expanded form. It's important to apply the right algebraic rules, like distribution or squaring binomials.
🎯 Exam Tip: Always remember the basic algebraic identities such as \( (a+b)^2 \) and \( (a-b)^2 \), and the distributive property. They are essential for simplifying expressions correctly.