Factorisation — NCERT Solutions, AI Tutor & Practice

Step 1: Write each term as a product of factors.

$12x = 2 \times 2 \times 3 \times x$

$36 = 2 \times 2 \times 3 \times 3$

Step 2: Pick the factors common to both.

Common factors: $2 \times 2 \times 3 = 12$

Answer: The common factor is $12$.

Step 1: Factorise each term.

$12a^2b = 3 \times 2 \times 2 \times a \times a \times b$

$15ab^2 = 3 \times 5 \times a \times b \times b$

Step 2: Find the common factors: $3 \times a \times b = 3ab$.

Step 3: Take out the common factor.

$12a^2b + 15ab^2 = 3ab(4a + 5b)$

Answer: $3ab(4a + 5b)$

Step 1: Rearrange and group the terms suitably.

$6xy - 4y + 6 - 9x = (6xy - 9x) + (-4y + 6)$

Step 2: Take common factors from each group.

$= 3x(2y - 3) - 2(2y - 3)$

Step 3: Take out the common binomial $(2y - 3)$.

$= (2y - 3)(3x - 2)$

Answer: $(2y - 3)(3x - 2)$

Step 1: Compare with the identity $a^2 + 2ab + b^2 = (a + b)^2$.

Here $a^2 = x^2 \Rightarrow a = x$ and $b^2 = 16 \Rightarrow b = 4$.

Step 2: Check the middle term.

$2ab = 2 \times x \times 4 = 8x$ ✓ (matches)

Step 3: Write the factorisation.

$x^2 + 8x + 16 = (x + 4)^2$

Answer: $(x + 4)^2$

Step 1: Recognise the identity $a^2 - b^2 = (a + b)(a - b)$.

$49p^2 = (7p)^2$ and $36 = (6)^2$.

Step 2: Apply the identity with $a = 7p$, $b = 6$.

$49p^2 - 36 = (7p + 6)(7p - 6)$

Answer: $(7p + 6)(7p - 6)$

Step 1: We need two numbers whose product is $+21$ and sum is $-10$.

The numbers are $-3$ and $-7$ (since $-3 \times -7 = 21$ and $-3 + (-7) = -10$).

Step 2: Split the middle term.

$x^2 - 10x + 21 = x^2 - 3x - 7x + 21$

Step 3: Group and factorise.

$= x(x - 3) - 7(x - 3) = (x - 3)(x - 7)$

Answer: $(x - 3)(x - 7)$

Step 1: Check for the identity $a^2 + 2ab + b^2 = (a+b)^2$.

$4x^2 = (2x)^2 \Rightarrow a = 2x$ and $9y^2 = (3y)^2 \Rightarrow b = 3y$.

Step 2: Verify the middle term.

$2ab = 2 \times 2x \times 3y = 12xy$ ✓

Step 3: Write the factorisation.

$4x^2 + 12xy + 9y^2 = (2x + 3y)^2$

Answer: $(2x + 3y)^2$

Step 1: Factorise the numerator.

$24(x^2yz + xy^2z + xyz^2) = 24 \times xyz(x + y + z)$

Step 2: Write the division.

$\dfrac{24 \, xyz(x + y + z)}{8xyz}$

Step 3: Cancel the common factors $8xyz$.

$= \dfrac{24}{8}(x + y + z) = 3(x + y + z)$

Answer: $3(x + y + z)$

Step 1: Take out the common factor from the numerator.

$3x^2 - 48 = 3(x^2 - 16)$

Step 2: Apply the difference of squares identity to $x^2 - 16$.

$x^2 - 16 = (x + 4)(x - 4)$

So $3x^2 - 48 = 3(x + 4)(x - 4)$.

Step 3: Divide by $(x - 4)$ and cancel.

$\dfrac{3(x + 4)(x - 4)}{(x - 4)} = 3(x + 4)$

Answer: $3(x + 4) = 3x + 12$

Step 1: Group the first three terms.

$a^2 - 2ab + b^2 - c^2 = (a^2 - 2ab + b^2) - c^2$

Step 2: The bracket is a perfect square.

$= (a - b)^2 - c^2$

Step 3: Apply difference of squares $X^2 - Y^2 = (X+Y)(X-Y)$ with $X = (a-b)$, $Y = c$.

$= (a - b + c)(a - b - c)$

Answer: $(a - b + c)(a - b - c)$

Step 1: Area = length × breadth, so factorise $x^2 + 7x + 10$.

Find two numbers with product $10$ and sum $7$: these are $2$ and $5$.

Step 2: Split the middle term.

$x^2 + 7x + 10 = x^2 + 2x + 5x + 10$

Step 3: Group and factorise.

$= x(x + 2) + 5(x + 2) = (x + 2)(x + 5)$

Step 4: Interpret the factors.

Length $= (x + 5)$ units and Breadth $= (x + 2)$ units.

Answer: Length $= (x+5)$, Breadth $= (x+2)$

Step 1: Identify the error. The student divided only the first term $5y$ by $5$ but forgot to divide the second term $20$ by $5$.

Step 2: Divide each term of the numerator by $5$.

$\dfrac{5y + 20}{5} = \dfrac{5y}{5} + \dfrac{20}{5} = y + 4$

Step 3 (verification by factorising):

$5y + 20 = 5(y + 4)$, so $\dfrac{5(y+4)}{5} = y + 4$.

Correct Answer: $y + 4$