CBSE Class 8 Mathematics Algebraic Expressions and Identities Assignment Set 06

Question 1. Using Suitable identity, evaluate
(a) \((x + 10)(x - 5)\)
Answer: Using the algebraic identity \((x + a)(x + b) = x^2 + (a + b)x + ab\), where \(a = 10\) and \(b = -5\):
\(= x^2 + (10 - 5)x + (10)(-5)\)
\(= x^2 + 5x - 50\)
In simple words: Multiply terms using the standard binomial identity for products of the form \((x + a)(x + b)\).

Exam Tip: Be careful with signs when substituting negative numbers into the identity formula.

Question 1. (b) \((x^2 + 5)(x^2 + 7)\)
Answer: Using the identity \((x + a)(x + b) = x^2 + (a + b)x + ab\), substituting \(x^2\) in place of \(x\), \(a = 5\), and \(b = 7\):
\(= (x^2)^2 + (5 + 7)(x^2) + (5)(7)\)
\(= x^4 + 12x^2 + 35\)
In simple words: Treat \(x^2\) as a single variable to apply the standard product identity directly.

Exam Tip: Remember that \((x^2)^2 = x^4\) due to exponent multiplication rules \((x^m)^n = x^{mn}\).

Question 1. (c) \((x - 3)(x - 4)\)
Answer: Using the identity \((x + a)(x + b) = x^2 + (a + b)x + ab\), with \(a = -3\) and \(b = -4\):
\(= x^2 + (-3 - 4)x + (-3)(-4)\)
\(= x^2 - 7x + 12\)
In simple words: The product of two negative numbers becomes positive.

Exam Tip: Double-check the middle term addition of negative integers \((-3 - 4 = -7)\).

Question 1. (d) \((2x + 3)(2x - 5)\)
Answer: Using the identity \((x + a)(x + b) = x^2 + (a + b)x + ab\), with \(x\) replaced by \(2x\), \(a = 3\), and \(b = -5\):
\(= (2x)^2 + (3 - 5)(2x) + (3)(-5)\)
\(= 4x^2 + (-2)(2x) - 15\)
\(= 4x^2 - 4x - 15\)
In simple words: Square the entire first term \((2x)^2\) to get \(4x^2\).

Exam Tip: Do not forget to square the coefficient 2 along with the variable \(x\).

Question 2. (a) \((2x + 3y)^2\)
Answer: Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), with \(a = 2x\) and \(b = 3y\):
\(= (2x)^2 + 2(2x)(3y) + (3y)^2\)
\(= 4x^2 + 12xy + 9y^2\)
In simple words: Expand the square of a binomial using the first term squared plus twice the product plus the second term squared.

Exam Tip: Ensure the middle term includes both coefficients multiplied together: \(2 \times 2 \times 3 = 12\).

Question 2. (b) \((3x - 2y)^2\)
Answer: Using the identity \((a - b)^2 = a^2 - 2ab + b^2\), with \(a = 3x\) and \(b = 2y\):
\(= (3x)^2 - 2(3x)(2y) + (2y)^2\)
\(= 9x^2 - 12xy + 4y^2\)
In simple words: The middle term is negative because of the subtraction sign in the binomial.

Question 2. (c) \((2x^2 - 3)^2\)
Answer: Using the identity \((a - b)^2 = a^2 - 2ab + b^2\), with \(a = 2x^2\) and \(b = 3\):
\(= (2x^2)^2 - 2(2x^2)(3) + (3)^2\)
\(= 4x^4 - 12x^2 + 9\)
In simple words: Apply exponent rules when squaring \(2x^2\) to obtain \(4x^4\).

Exam Tip: Keep track of variable powers carefully during multiplication.

Question 2. (d) \((3x^2 + 5)^2\)
Answer: Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), with \(a = 3x^2\) and \(b = 5\):
\(= (3x^2)^2 + 2(3x^2)(5) + (5)^2\)
\(= 9x^4 + 30x^2 + 25\)
In simple words: Square the first term, find twice the product of both terms, and square the last term.

Exam Tip: Double check multiplication: \(2 \times 3 \times 5 = 30\).

Question 3. (a) \((5x + 2y)(5x - 2y)\)
Answer: Using the identity \((a + b)(a - b) = a^2 - b^2\), with \(a = 5x\) and \(b = 2y\):
\(= (5x)^2 - (2y)^2\)
\(= 25x^2 - 4y^2\)
In simple words: The product of the sum and difference of two terms is the difference of their squares.

Exam Tip: This is one of the most frequently used identities in factorization.

Question 3. (b) \((x + 2y)(x - 2y)\)
Answer: Using the identity \((a + b)(a - b) = a^2 - b^2\), with \(a = x\) and \(b = 2y\):
\(= (x)^2 - (2y)^2\)
\(= x^2 - 4y^2\)
In simple words: Square the first term and subtract the square of the second term.

Exam Tip: Ensure both parts of the second term \((2y)\) are squared properly.

Question 3. (c) \((4x - 3y)(4x + 3y)\)
Answer: Using the identity \((a - b)(a + b) = a^2 - b^2\), with \(a = 4x\) and \(b = 3y\):
\(= (4x)^2 - (3y)^2\)
\(= 16x^2 - 9y^2\)
In simple words: Order of factors in multiplication does not change the result.

Exam Tip: \(4^2 = 16\) and \(3^2 = 9\).

Question 3. (d) \((\frac{5}{6}x - \frac{2}{3}y)(\frac{5}{6}x + \frac{2}{3}y)\)
Answer: Using the identity \((a - b)(a + b) = a^2 - b^2\), with \(a = \frac{5}{6}x\) and \(b = \frac{2}{3}y\):
\(= (\frac{5}{6}x)^2 - (\frac{2}{3}y)^2\)
\(= \frac{25}{36}x^2 - \frac{4}{9}y^2\)
In simple words: Square both the numerator and the denominator of the fractions.

Exam Tip: Squaring fractions means squaring top and bottom numbers independently: \(5^2=25\), \(6^2=36\).

Question 4. (a) If \(x + \frac{1}{x} = 2\), Find \(x^2 + \frac{1}{x^2}\)
Answer: Given \(x + \frac{1}{x} = 2\)
Squaring both sides:
\((x + \frac{1}{x})^2 = (2)^2\)
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(x^2 + 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 4\)
\(x^2 + 2 + \frac{1}{x^2} = 4\)
\(x^2 + \frac{1}{x^2} = 4 - 2\)
\(x^2 + \frac{1}{x^2} = 2\)
In simple words: Square the given equation and simplify the middle term where \(x\) cancels out with \(\frac{1}{x}\).

Exam Tip: Always remember that \(x \times \frac{1}{x} = 1\).

Question 4. (b) If \(x - \frac{1}{x} = 2\), Find \(x^2 + \frac{1}{x^2}\)
Answer: Given \(x - \frac{1}{x} = 2\)
Squaring both sides:
\((x - \frac{1}{x})^2 = (2)^2\)
Using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(x^2 - 2(x)(\frac{1}{x}) + (\frac{1}{x})^2 = 4\)
\(x^2 - 2 + \frac{1}{x^2} = 4\)
\(x^2 + \frac{1}{x^2} = 4 + 2\)
\(x^2 + \frac{1}{x^2} = 6\)
In simple words: Square both sides and transpose the negative constant to the right side.

Exam Tip: Notice that even with a minus sign in the given equation, the required expression asks for \(x^2 + \frac{1}{x^2}\) which remains positive.

Question 5. Find the product \(( \frac{3}{4}x^2 + \frac{2}{5}y^2 )^2\)
Answer: Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), with \(a = \frac{3}{4}x^2\) and \(b = \frac{2}{5}y^2\):
\(= (\frac{3}{4}x^2)^2 + 2(\frac{3}{4}x^2)(\frac{2}{5}y^2) + (\frac{2}{5}y^2)^2\)
\(= \frac{9}{16}x^4 + 2 \times \frac{6}{20} x^2 y^2 + \frac{4}{25} y^4\)
\(= \frac{9}{16}x^4 + \frac{12}{20}x^2y^2 + \frac{4}{25}y^4\)
\(= \frac{9}{16}x^4 + \frac{3}{5}x^2y^2 + \frac{4}{25}y^4\)
In simple words: Expand using the square of binomial formula and simplify the fractional coefficients.

Exam Tip: Simplify fraction coefficients completely in the final step.

Question 6. Find continued product of \((x + y)(x - y)(x^2 + y^2)(x^4 + y^4)\)
Answer: Group the first two terms and apply the identity \((a + b)(a - b) = a^2 - b^2\):
\(= [ (x + y)(x - y) ] (x^2 + y^2)(x^4 + y^4)\)
\(= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)\)
Now apply the identity again on the first two terms: \((x^2 - y^2)(x^2 + y^2) = (x^2)^2 - (y^2)^2 = x^4 - y^4\)
\(= (x^4 - y^4)(x^4 + y^4)\)
Apply the identity one more time: \((x^4 - y^4)(x^4 + y^4) = (x^4)^2 - (y^4)^2 = x^8 - y^8\)
In simple words: Repeatedly apply the difference of squares identity from left to right.

Exam Tip: Step-by-step pairing makes complex continued products very simple to solve.

Question 7. If \((x + y) = 11\) and \(xy = 30\), Find the value of \(x^2 + y^2\)
Answer: Consider the identity \((x + y)^2 = x^2 + 2xy + y^2\)
Substitute the given values \(x + y = 11\) and \(xy = 30\):
\((11)^2 = x^2 + 2(30) + y^2\)
\(121 = x^2 + 60 + y^2\)
\(x^2 + y^2 = 121 - 60\)
\(x^2 + y^2 = 61\)
In simple words: Square the sum equation and substitute the known product value to find the sum of squares.

Exam Tip: Memorize the relation connecting \((x+y)^2\), \((x^2+y^2)\), and \(xy\).

Question 8. Find the product of \((x - 1)(x + 1)(x^2 + 1)\)
Answer: Combine the first two terms using the identity \((a - b)(a + b) = a^2 - b^2\):
\(= [ (x - 1)(x + 1) ] (x^2 + 1)\)
\(= (x^2 - 1)(x^2 + 1)\)
Now apply the identity again with \(a = x^2\) and \(b = 1\):
\(= (x^2)^2 - (1)^2\)
\(= x^4 - 1\)
In simple words: Group the conjugate pair first, then apply the difference of squares identity again.

Exam Tip: \(1^2\) is always equal to 1.

Question 9. Find product of \((3x + 2y)(3x - 2y)(9x^2 + 4y^2)\)
Answer: Apply the identity \((a + b)(a - b) = a^2 - b^2\) on the first two terms:l
\(= [ (3x + 2y)(3x - 2y) ] (9x^2 + 4y^2)\)
\(= [ (3x)^2 - (2y)^2 ] (9x^2 + 4y^2)\)
\(= (9x^2 - 4y^2)(9x^2 + 4y^2)\)
Now apply the difference of squares identity again with \(a = 9x^2\) and \(b = 4y^2\):
\(= (9x^2)^2 - (4y^2)^2\)
\(= 81x^4 - 16y^4\)
In simple words: Simplify the first binomial product first, then use the difference of squares formula for the remaining binomials.

Exam Tip: Remember to square both the coefficient and variables: \((9x^2)^2 = 81x^4\).

Question 10. Simplify, \((\frac{2}{5}x^2 + \frac{7}{9}y^2)^2 - (\frac{2}{5}x^2 - \frac{7}{9}y^2)^2\)
Answer: Let \(a = \frac{2}{5}x^2\) and \(b = \frac{7}{9}y^2\). The expression is of the form \((a + b)^2 - (a - b)^2\).
Using the standard identity \((a + b)^2 - (a - b)^2 = 4ab\):
\(= 4 (\frac{2}{5}x^2)(\frac{7}{9}y^2)\)
Multiply the numerators and denominators:
Numerator = \(4 \times 2 \times 7 = 56\)
Denominator = \(5 \times 9 = 45\)
Variables = \(x^2 y^2\)
\(= \frac{56}{45}x^2y^2\)
In simple words: Use the special expansion identity \((a+b)^2 - (a-b)^2 = 4ab\) to save time and effort.

Exam Tip: Recognizing standard algebraic expansion shortcuts like \(4ab\) prevents long, error-prone polynomial expansions.