गुणनखंडन (Factorisation)

A. What is Factorisation?

Factorisation is the reverse process of multiplication. It involves expressing a given algebraic polynomial as the product of two or more irreducible factors:

• An Irreducible Factor is a factor that cannot be factored further into simpler algebraic expressions (e.g., in $3x(x + 2)$, the irreducible factors are $3, x,$ and $(x + 2)$).

B. Method 1: Taking out the Highest Common Factor (HCF):

1. Find the numerical HCF of all the coefficients.
2. Find the lowest power of each common variable present in every term.
3. Write the product of the numerical HCF and common variables outside parentheses, and divide each term by this common factor.
4. Example: Factorise $12x^3y^2 - 18x^2y^3 + 24x^4y$:
   • Numerical HCF of 12, 18, 24 is 6.
   • Lowest powers of common variables: $x^2$ and $y$. Overall $\text{HCF} = \mathbf{6x^2y}$.
   • Divide each term: $\frac{12x^3y^2}{6x^2y} = 2xy; \quad \frac{-18x^2y^3}{6x^2y} = -3y^2; \quad \frac{24x^4y}{6x^2y} = 4x^2$.
   • $$\mathbf{12x^3y^2 - 18x^2y^3 + 24x^4y = 6x^2y(2xy - 3y^2 + 4x^2)}$$

A. The Four-Term Grouping Algorithm:

When an expression has four terms with no single factor common to all terms, group the terms into pairs that share a common factor:

1. Group terms into two pairs: $(T_1 + T_2) + (T_3 + T_4)$. Sometimes terms must be rearranged first!
2. Take out the common factor from each pair to reveal an identical common binomial bracket.
3. Factor out the common binomial bracket as the final product.
4. Example: Factorise $ab + bc + ax + cx$: $$(ab + bc) + (ax + cx) = b(a + c) + x(a + c) = \mathbf{(a + c)(b + x)}$$
5. Sign-Trap Example: Factorise $x^2 - x(a + 2b) + 2ab$: $$x^2 - ax - 2bx + 2ab = x(x - a) - 2b(x - a) = \mathbf{(x - a)(x - 2b)}$$ *(Notice that taking out $-2b$ flips $+2ab$ to $-a$!)*.

A. Difference of Two Squares:

• Iterated / Nested Difference of Squares: $$x^4 - 81 = (x^2)^2 - 9^2 = (x^2 - 9)(x^2 + 9) = \mathbf{(x - 3)(x + 3)(x^2 + 9)}$$ *(Always check if the resulting factor can be factored again!)*
• Grouped Difference of Squares: $$x^2 - y^2 - 4x + 4 = (x^2 - 4x + 4) - y^2 = (x - 2)^2 - y^2 = \mathbf{(x - 2 - y)(x - 2 + y)}$$

B. Perfect Square Trinomials:

Example: $4x^2 - 12xy + 9y^2 = (2x)^2 - 2(2x)(3y) + (3y)^2 = \mathbf{(2x - 3y)^2}$.

A. General Trinomial Form: $ax^2 + bx + c$:

To factorise $ax^2 + bx + c$, find two numbers $p$ and $q$ such that:

1. Replace the middle term $bx$ with $px + qx$.
2. Factorise the resulting four-term expression by grouping in pairs.

B. Master Worked Examples:

• Case 1 ($a = 1$, Positive constant): Factorise $x^2 + 9x + 20$:

  Product $p \cdot q = 20$, Sum $p + q = 9$. Factors are $4$ and $5$ ($4 \times 5 = 20, 4 + 5 = 9$).

  $$x^2 + 4x + 5x + 20 = x(x + 4) + 5(x + 4) = \mathbf{(x + 4)(x + 5)}$$
• Case 2 ($a > 1$, Negative middle term): Factorise $6x^2 - 17x + 12$:

  Product $ac = 6 \times 12 = 72$. Sum $b = -17$. Two numbers are $-9$ and $-8$ ($(-9)(-8) = 72, -9 - 8 = -17$).

  $$6x^2 - 9x - 8x + 12 = 3x(2x - 3) - 4(2x - 3) = \mathbf{(2x - 3)(3x - 4)}$$
• Case 3 (Negative constant): Factorise $2x^2 - x - 15$:

  Product $ac = 2 \times (-15) = -30$. Sum $b = -1$. Factors are $-6$ and $+5$ ($(-6)(5) = -30, -6 + 5 = -1$).

  $$2x^2 - 6x + 5x - 15 = 2x(x - 3) + 5(x - 3) = \mathbf{(x - 3)(2x + 5)}$$