Factorization - 2 | Lesson Note

⬜ F) Factorization of Trinomials | (a) Completing the Square

Let us consider the algebraic expression:

$$a^2 + 2ab + b^2$$

        💡 We can clearly see that the first term $a^2$ and the last term $b^2$ are both in square form.

        So here we can use the completing the square method to factorize.

❓ How does it work?

\phantom{=}\ a^2 + 2ab + b^2$ $= (a)^2 + 2 \cdot a \cdot b + (b)^2$ $= (a + b)^2$ $= (a + b)(a + b)

📌 Example (a): Factorize $x^2 - 10x + 25$

First, write the first term and last term in square form, as $(x)^2$ and $(5)^2$:

= (x)^2 - 2 \cdot x \cdot 5 + (5)^2$ This is in the form $a^2 - 2ab + b^2$, which is equivalent to $(a-b)^2$ $= (x - 5)^2$ $= (x - 5)(x - 5)

📌 Example (b): Factorize $9x^2 - 30xy + 25y^2$

First, write the first term $9x^2$ and last term $25y^2$ in square form:

9x^2 = (3x)^2 \qquad \text{and} \qquad 25y^2 = (5y)^2$ $= (3x)^2 - 2 \cdot 3x \cdot 5y + (5y)^2$ This is in the form $a^2 - 2ab + b^2$, which is equivalent to $(a - b)^2$ $= (3x - 5y)^2$ $= (3x - 5y)(3x - 5y)

📌 Example (c): Factorize $4x^2 + xy + \dfrac{y^2}{16}$

First, write the first term $4x^2$ and last term $\dfrac{y^2}{16}$ in square form:

4x^2 = (2x)^2 \qquad \text{and} \qquad \dfrac{y^2}{16} = \left(\dfrac{y}{4}\right)^2

Now build the expression step by step:

= (2x)^2 \phantom{{}+ 2 \cdot 2x \cdot \tfrac{y}{4}} + \left(\dfrac{y}{4}\right)^2$ $\downarrow$   (we need the middle term $2ab$) $= (2x)^2 + 2 \phantom{{}\cdot 2x \cdot \tfrac{y}{4}} + \left(\dfrac{y}{4}\right)^2$ $\downarrow$ $= (2x)^2 + 2 \cdot 2x \phantom{{}\cdot \tfrac{y}{4}} + \left(\dfrac{y}{4}\right)^2$ $\downarrow$ $= (2x)^2 + 2 \cdot 2x \cdot \dfrac{y}{4} + \left(\dfrac{y}{4}\right)^2

This is in the form $a^2 + 2ab + b^2$, which is equivalent to $(a + b)^2$

= \left(2x + \dfrac{y}{4}\right)^2$ $= \left(2x + \dfrac{y}{4}\right)\left(2x + \dfrac{y}{4}\right)

✏️ Extra Questions — What must be filled in the gap so that the expression is a perfect square?

(a) $x^2 + \text{\_\_\_\_} + 4$

Solution:

(x)^2 + \text{\_\_\_\_} + (2)^2 = (a)^2 + \underbrace{2ab}_{\text{gap}} + (b)^2$ $a = x, \quad b = 2$ Gap $= 2ab = 2 \times x \times 2 = 4x

∴ The missing term is $\mathbf{4x}$, giving the perfect square $(x + 2)^2$.

(b) $9x^2 - \text{\_\_\_\_} + 16$

Solution:

(3x)^2 - \text{\_\_\_\_} + (4)^2 = (a)^2 - \underbrace{2ab}_{\text{gap}} + (b)^2$ $a = 3x, \quad b = 4$ Gap $= 2ab = 2 \times 3x \times 4 = 24x

∴ The missing term is $\mathbf{24x}$, giving the perfect square $(3x - 4)^2$.

🔬 Activity — A Trinomial of the Form $ax^2 + bx + c$ Always Represents the Area of a Rectangle

Given materials provided to students:

One tile of area $A_1 = x^2$ (a square of side $x$)
Three tiles of area $A_2 = A_3 = A_4 = 1 \cdot x$ (rectangles of dimensions $1 \times x$)
Two unit tiles: $A_5 = 1,\; A_6 = 1$

Students are asked to make a flawless rectangle from the given tiles.

(After several attempts, students made this rectangle:)

        Observation:
        Length $(\ell) = x + 1 + 1 = x + 2$
Breadth $(b) = x + 1$
Area $(A) = \ell \times b$

      

A_1 + A_2 + A_3 + A_4 + A_5 + A_6 = (x+2)(x+1)$ $x^2 + 1\cdot x + 1\cdot x + 1\cdot x + 1 + 1 = (x+2)(x+1)$ $x^2 + 3x + 2 = (x+2)(x+1)

∴ $(x+2)$ and $(x+1)$ are factors of $x^2 + 3x + 2$.

        Conclusion: A trinomial of the form $ax^2 + bx + c$ always represents the area of a rectangle.
      

✂️ (B) Factorization of $ax^2 + bx + c$ using the AC Method

Methods (Steps):

Identify $a$ and $c$ from the given trinomial, where $a$ is the coefficient of $x^2$ and $c$ is the constant.
Find the product $P = ac$.
Case (i) — If $P$ is positive: find two numbers $n_1$ and $n_2$ such that $n_1 + n_2 = b \quad \text{(the coefficient of } x\text{, i.e. the middle term)}$ where $n_1$ and $n_2$ are parts of $b$.
Case (ii) — If $P$ is negative: find two numbers $n_1$ and $n_2$ such that $$$n_1 - n_2 = b$$$

Real Examples:

📌 Example (i): Factorize $3x^2 + 10x + 8$

Here, $a = 3,\; b = 10,\; c = 8$

$P = ac = 3 \times 8 = 24$ (positive) $\Rightarrow$ find $n_1 + n_2 = 10$

Find two numbers whose product is 24 and sum is 10:

$P = 24 \;\;(n_1 \times n_2 = 24)$	$n_1 + n_2 = 10$
$1 \times 24 = 24$	$24 + 1 = 25$ (greater than 10)
$2 \times 12 = 24$	$12 + 2 = 14$ (getting close)
$3 \times 8 = 24$	$8 + 3 = 11$ (almost there)
$4 \times 6 = 24$	$6 + 4 = 10$ ✓ (exactly there!)

∴ 6 and 4 are the required two numbers.

Split the middle term and factorize (take two terms at a time):

$= 3x^2 + (6 + 4)x + 8$ $= 3x^2 + 6x + 4x + 8$ $= 3 \cdot x \cdot x + 2 \cdot 3 \cdot x + 2 \cdot 2 \cdot x + 2 \cdot 2 \cdot 2$ $= 3x(x + 2) + 2 \cdot 2(x + 2)$ $= 3x(x + 2) + 4(x + 2)$ $= (x + 2)(3x + 4)$   (taking common factor)

📌 Example (ii): Factorize $2x^2 - x - 15$

Here, $a = 2,\; b = -1,\; c = -15$

$P = ac = 2 \times (-15) = -30$ (negative) $\Rightarrow$ find $n_1 - n_2 = -1$

Find two numbers $n_1$ and $n_2$ whose product is 30 and difference is $-1$:

$P = 30 \;\;(n_1 \times n_2 = 30)$	$n_1 - n_2 = -1$
$1 \times 30 = 30$	$1 - 30 = -29$ (very far)
$2 \times 15 = 30$	$2 - 15 = -13$ (somewhat close)
$3 \times 10 = 30$	$3 - 10 = -7$ (getting close)
$5 \times 6 = 30$	$5 - 6 = -1$ ✓ (oh! we got there)

∴ 5 and 6 are the required numbers.

Split the middle term and factorize:

= 2x^2 + (5 - 6)x - 15$ $= 2x^2 + 5x - 6x - 15$ $= 2 \cdot x \cdot x + 5 \cdot x - 2 \cdot 3 \cdot x - 3 \cdot 5$ $= x(2x + 5) - 3(2x + 5)$ $= (2x + 5)(x - 3)

            💬 Sign note for students: No need to worry about what sign to write — once we see the factor $(2x + 5)$ from the first two terms, students must understand that from the last two terms we should also get the same factor $(2x + 5)$. So the sign will be " + ".
          

⭐ Example (iii) — Very Important: For what value of $a$, will $a^2 - 7a + 12$ be zero? Justify.

Solution: According to the question (ATQ):

a^2 - 7a + 12 = 0$ $a^2 - (4a + 3a) + 12 = 0$   (split: $4 + 3 = 7$ and $4 \times 3 = 12$) $a^2 - 4a - 3a + 12 = 0$ $a \cdot a - 4 \cdot a - 3 \cdot a + 3 \cdot 4 = 0$ $a(a - 4) - 3(a - 4) = 0$ $(a - 3)(a - 4) = 0

              Either:

              $a - 3 = 0$

              $\therefore\; a = 3$

              Or:

              $a - 4 = 0$

              $\therefore\; a = 4$

Checking:

When $a = 3$	When $a = 4$
$= 3^2 - 7 \times 3 + 12$	$= 4^2 - 7 \times 4 + 12$
$= 9 - 21 + 12$	$= 16 - 28 + 12$
$= 21 - 21$	$= 28 - 28$
$= 0\quad$ ✓	$= 0\quad$ ✓

∴ When $a = 3$ and $a = 4$, the value of the expression $a^2 - 7a + 12$ is zero.