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Multiplication and Standard Identities Quiz - CBSE Class 8 Mathematics

Free practice quiz: Multiplication and Standard Identities (CBSE Class 8 Mathematics). 14 multiple-choice questions with answers and explanations. Play online, no sign-up needed.

This CBSE Class 8 Mathematics practice quiz has 14 multiple-choice questions on Multiplication and Standard Identities, and 14 of them come with an explanation. Try the 5 sample questions below (tap an option to check yourself), then play the full quiz online. No sign-up needed.

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Multiplying Polynomials Term by Term

To multiply two expressions, multiply every term of the first by every term of the second.
Then add like terms together to get the simplest answer.

Monomial times binomial

\[p(q + r) = pq + pr\]

Binomial times binomial

\[(p + q)(r + s) = pr + ps + qr + qs\]

Common mistake: Forgetting to multiply one pair of terms, or getting a sign wrong when multiplying negative terms.

Standard Square Identities

These identities are shortcuts for squaring or multiplying binomials without full multiplication.
They also help with quick arithmetic. Write a number as a round number plus or minus a small part.

Square of a sum

\[(p + q)^2 = p^2 + 2pq + q^2\]

Square of a difference

\[(p - q)^2 = p^2 - 2pq + q^2\]

Sum times difference

\[(p + q)(p - q) = p^2 - q^2\]

Two binomials with a common term

\[(y + m)(y + n) = y^2 + (m + n)y + mn\]

Common mistake: Writing (p + q)^2 = p^2 + q^2 and forgetting the middle term 2pq.

Finding Values Using Identities

If the sum and product of two quantities are known, rearrange an identity to find the sum of their squares.
The same idea works for a quantity and its reciprocal, because their product is 1.

Sum of squares from sum and product

\[p^2 + q^2 = (p + q)^2 - 2pq\]

Difference squared from sum and product

\[(p - q)^2 = (p + q)^2 - 4pq\]

Reciprocal form with sum

\[m^2 + \frac{1}{m^2} = \left(m + \frac{1}{m}\right)^2 - 2\]

Reciprocal form with difference

\[m^2 + \frac{1}{m^2} = \left(m - \frac{1}{m}\right)^2 + 2\]

Common mistake: Using -2 with the difference form. For (m - 1/m), you must add 2, not subtract it.

Three-Term Squares and Cube Identities

Squaring three terms gives every square plus twice every pair product.
Cube identities expand (p + q)^3 and (p - q)^3 directly. Watch the alternating signs in the difference cube.

Square of three terms

\[(p + q + r)^2 = p^2 + q^2 + r^2 + 2pq + 2qr + 2rp\]

Cube of a sum

\[(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3\]

Cube of a difference

\[(p - q)^3 = p^3 - 3p^2q + 3pq^2 - q^3\]

Sum of three cubes

\[p^3 + q^3 + r^3 - 3pqr = (p + q + r)(p^2 + q^2 + r^2 - pq - qr - rp)\]

Common mistake: Writing (p + q)^3 = p^3 + q^3, or getting the signs wrong in the expansion of (p - q)^3.

Exam tips

For quick squares, split the number near a round value. For example, 48 = 50 - 2, then use an identity.
When factorising, apply the difference of squares again if a factor is still a difference of squares.
Check your expansion by substituting a small value, like 1 or 2, on both sides.

Beta: made with AI and checked by a second AI. A teacher has not reviewed it yet.

Sample questions

Expand (x + 3)(x - 3)
Show answer
Answer: A. x² - 9
This is a difference of squares. (a + b)(a - b) = a² - b². Therefore, (x + 3)(x - 3) = x² - 3² = x² - 9.
Find the value of (99)² using a suitable identity.
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Answer: A. 9801
We can write 99 as (100 - 1). Then (99)² = (100 - 1)² = 100² - 2(100)(1) + 1² = 10000 - 200 + 1 = 9801. This uses the identity (a - b)² = a² - 2ab + b².
Simplify (x + y)(x - y) + (x + y)²
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Answer: A. 2x² + 2xy
Expand each part: (x + y)(x - y) = x² - y² and (x + y)² = x² + 2xy + y². Adding them, we get x² - y² + x² + 2xy + y² = 2x² + 2xy.
If a + b + c = 0, prove that a³ + b³ + c³ = 3abc
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Answer: A. True
This is a standard identity. Start with the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). If a + b + c = 0, the right side becomes 0, implying a³ + b³ + c³ = 3abc.
Simplify (a+b)(a-b) - (a-b)^2
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Answer: A. 2b(a-b)
First expand: (a+b)(a-b) = a^2 - b^2 and (a-b)^2 = a^2 - 2ab + b^2. Subtracting the second from the first gives (a^2 - b^2) - (a^2 - 2ab + b^2) = 2ab - 2b^2 = 2b(a-b)

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