Factorisation — Statement I-II Type Questions

Statement I: The HCF of $6x^2y^3z$ and $12x^2z^3$ is $6x^2yz$

Statement II: HCF of literal coefficients = product of each common literal.

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

HCF of numerical coefficients = HCF of 6 and 12 = 6

The literal $y$ occurs in $6x^2y^3z$ but not in $12x^2z^3$, so it is not a common literal.

HCF of literal coefficients = HCF of $x^2y^3z$ and $x^2z^3$ = $x^2z$

HCF of the given monomials = 6 × $x^2z$ = $6x^2z$, which is not $6x^2yz$.

So, Statement I is false.

HCF of literal coefficients = product of each common literal raised to the lowest power, and not merely the product of each common literal.

So, Statement II is false.

Hence, option 4 is the correct option.

Statement I: The factors of $3abc^2 + ab$ are $(3c^2 + 1)$ and $(ab + 1)$

Statement II: A polynomial with even number of terms can be factorised by grouping.

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

HCF of $3abc^2$ and $ab$ is $ab$

Divide each term by $ab$ and keep $ab$ outside the bracket.

$\Rightarrow 3abc^2 + ab \\[1em] \Rightarrow ab(3c^2 + 1).$

So, the factors are $ab$ and $(3c^2 + 1)$, and not $(ab + 1)$.

So, Statement I is false.

Factorisation by grouping is possible only if the given polynomial contains an even number of terms.

Thus,

A polynomial with even number of terms can be factorised by grouping.

So, Statement II is true.

Hence, option 2 is the correct option.

Statement I: The factors of $(x + 5)^2 - 1$ are $(x + 6)$ and $(x + 4)$

Statement II: $(a + b)(a - b) = a^2 - b^2$

1. Statement I is true but statement II is false.

2. Statement I is false but statement II is true.

3. Both Statement I and statement II are true.

4. Both Statement I and statement II are false.

Answer

Factorising,

⇒ $(x + 5)^2 - 1$

⇒ $(x + 5)^2 - (1)^2$

Using the identity,

[∵ $(x^2 - y^2) = (x + y)(x - y)$]

⇒ $((x + 5) + 1)((x + 5) - 1)$

⇒ $(x + 6)(x + 4)$

So, Statement I is true.

$(a + b)(a - b) = a^2 - b^2$ is a standard algebraic identity.

So, Statement II is true.

Hence, option 3 is the correct option.