Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.6

Question 1. Factorise the following.
(i) \( x^2 + 10x + 24 \)
(ii) \( z^2 + 4z - 12 \)
(iii) \( p^2 - 6p - 16 \)
(iv) \( t^2 + 72 - 17t \)
(v) \( y^2 - 16y - 80 \)
(vi) \( a^2 + 10a - 600 \)
Answer:
(i) We need to find two numbers that multiply to 24 and add up to 10. These numbers are 6 and 4.
\( x^2 + 10x + 24 = x^2 + 6x + 4x + 24 \)
\( = x(x + 6) + 4(x + 6) \)
\( = (x + 6)(x + 4) \)
(ii) We need two numbers that multiply to -12 and add up to 4. These numbers are 6 and -2.
\( z^2 + 4z - 12 = z^2 + 6z - 2z - 12 \)
\( = z(z + 6) - 2(z + 6) \)
\( = (z + 6)(z - 2) \)
(iii) We need two numbers that multiply to -16 and add up to -6. These numbers are -8 and 2.
\( p^2 - 6p - 16 = p^2 - 8p + 2p - 16 \)
\( = p(p - 8) + 2(p - 8) \)
\( = (p - 8)(p + 2) \)
(iv) First, rearrange the terms in order of powers: \( t^2 - 17t + 72 \). We need two numbers that multiply to 72 and add up to -17. These numbers are -9 and -8.
\( t^2 - 17t + 72 = t^2 - 9t - 8t + 72 \)
\( = t(t - 9) - 8(t - 9) \)
\( = (t - 9)(t - 8) \)
(v) We need two numbers that multiply to -80 and add up to -16. These numbers are -20 and 4.
\( y^2 - 16y - 80 = y^2 - 20y + 4y - 80 \)
\( = y(y - 20) + 4(y - 20) \)
\( = (y - 20)(y + 4) \)
(vi) We need two numbers that multiply to -600 and add up to 10. These numbers are 30 and -20.
\( a^2 + 10a - 600 = a^2 + 30a - 20a - 600 \)
\( = a(a + 30) - 20(a + 30) \)
\( = (a + 30)(a - 20) \)
In simple words: To factorise a quadratic expression like \( x^2 + bx + c \), find two numbers that multiply to \( c \) and add up to \( b \). Then, rewrite the middle term \( bx \) using these two numbers and group the terms to find the common factors.

🎯 Exam Tip: Always check your factoring by multiplying the factors back together to ensure you get the original expression. This helps catch any mistakes.

Question 2. Factorise the following.
(i) \( 2a^2 + 9a + 10 \)
(ii) \( 5x^2 - 29xy - 42y^2 \)
(iii) \( 9 - 18x + 8x^2 \)
(iv) \( 6x^2 + 16xy + 8y^2 \)
(v) \( 3x^2(9y^2 + 12y + 4) \)
(vi) \( (a + b)^2 + 9(a + b) + 18 \)
Answer:
(i) For \( 2a^2 + 9a + 10 \), we multiply the first and last coefficients: \( 2 \times 10 = 20 \). We need two numbers that multiply to 20 and add up to 9. These are 5 and 4.
\( 2a^2 + 9a + 10 = 2a^2 + 5a + 4a + 10 \)
\( = a(2a + 5) + 2(2a + 5) \)
\( = (2a + 5)(a + 2) \)
(ii) For \( 5x^2 - 29xy - 42y^2 \), we multiply the coefficients of \( x^2 \) and \( y^2 \): \( 5 \times (-42) = -210 \). We need two numbers that multiply to -210 and add up to -29. These are -35 and 6.
\( 5x^2 - 29xy - 42y^2 = 5x^2 - 35xy + 6xy - 42y^2 \)
\( = 5x(x - 7y) + 6y(x - 7y) \)
\( = (x - 7y)(5x + 6y) \)
(iii) First, rearrange the terms in descending powers of \( x \): \( 8x^2 - 18x + 9 \). We multiply the first and last coefficients: \( 8 \times 9 = 72 \). We need two numbers that multiply to 72 and add up to -18. These are -12 and -6.
\( 8x^2 - 18x + 9 = 8x^2 - 12x - 6x + 9 \)
\( = 4x(2x - 3) - 3(2x - 3) \)
\( = (2x - 3)(4x - 3) \)
(iv) For \( 6x^2 + 16xy + 8y^2 \), we multiply the coefficients of \( x^2 \) and \( y^2 \): \( 6 \times 8 = 48 \). We need two numbers that multiply to 48 and add up to 16. These are 12 and 4.
\( 6x^2 + 16xy + 8y^2 = 6x^2 + 12xy + 4xy + 8y^2 \)
\( = 6x(x + 2y) + 4y(x + 2y) \)
\( = (x + 2y)(6x + 4y) \)
We can also take out a common factor of 2 from \( (6x + 4y) \):
\( = (x + 2y) \cdot 2(3x + 2y) \)
\( = 2(x + 2y)(3x + 2y) \)
(v) For \( 3x^2(9y^2 + 12y + 4) \), we look at the expression inside the parenthesis. This expression, \( 9y^2 + 12y + 4 \), is a perfect square trinomial.
We can write \( 9y^2 \) as \( (3y)^2 \) and \( 4 \) as \( (2)^2 \). The middle term \( 12y \) is \( 2 \times 3y \times 2 \).
So, \( 9y^2 + 12y + 4 = (3y + 2)^2 \).
Therefore, \( 3x^2(9y^2 + 12y + 4) = 3x^2(3y + 2)^2 \).
(vi) For \( (a + b)^2 + 9(a + b) + 18 \), let's substitute \( x = (a + b) \). The expression becomes \( x^2 + 9x + 18 \).
We need two numbers that multiply to 18 and add up to 9. These numbers are 6 and 3.
\( x^2 + 9x + 18 = x^2 + 6x + 3x + 18 \)
\( = x(x + 6) + 3(x + 6) \)
\( = (x + 6)(x + 3) \)
Now, substitute back \( x = (a + b) \):
\( = (a + b + 6)(a + b + 3) \)
In simple words: When factoring expressions where the first term has a coefficient greater than 1, multiply the first and last coefficients together to find the product. Then find two numbers that fit this product and the middle term's sum. If the expression has a common part, like \( (a+b) \), you can temporarily replace it with a single letter to make factoring easier. Remember to put the original expression back in the end.

🎯 Exam Tip: For expressions with \( ax^2 + bx + c \) form, multiply \( a \cdot c \) to find the product for splitting the middle term. For expressions with common compound terms like \( (a+b) \), use substitution to simplify the factoring process.

Question 3. Factorise the following.
(i) \( (p - q)^2 - 6(p - q) - 16 \)
(ii) \( m^2 + 2mn - 24n^2 \)
(iii) \( \sqrt{5} a^2 + 2a - 3\sqrt{5} \)
(iv) \( a^4 - 3a^2 + 2 \)
(v) \( 8m^3 - 2m^2n - 15mn^2 \)
(vi) \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{2}{xy} \)
Answer:
(i) For \( (p - q)^2 - 6(p - q) - 16 \), let's substitute \( x = (p - q) \). The expression becomes \( x^2 - 6x - 16 \).
We need two numbers that multiply to -16 and add up to -6. These numbers are -8 and 2.
\( x^2 - 6x - 16 = x^2 - 8x + 2x - 16 \)
\( = x(x - 8) + 2(x - 8) \)
\( = (x - 8)(x + 2) \)
Now, substitute back \( x = (p - q) \):
\( = (p - q - 8)(p - q + 2) \)
(ii) For \( m^2 + 2mn - 24n^2 \), we need two numbers that multiply to -24 and add up to 2. These numbers are 6 and -4.
\( m^2 + 2mn - 24n^2 = m^2 + 6mn - 4mn - 24n^2 \)
\( = m(m + 6n) - 4n(m + 6n) \)
\( = (m + 6n)(m - 4n) \)
(iii) For \( \sqrt{5} a^2 + 2a - 3\sqrt{5} \), we multiply the coefficients of the first and last terms: \( \sqrt{5} \times (-3\sqrt{5}) = -3 \times 5 = -15 \). We need two numbers that multiply to -15 and add up to 2. These are 5 and -3.
\( \sqrt{5} a^2 + 2a - 3\sqrt{5} = \sqrt{5} a^2 + 5a - 3a - 3\sqrt{5} \)
\( = \sqrt{5}a(a + \sqrt{5}) - 3(a + \sqrt{5}) \)
\( = (a + \sqrt{5})(\sqrt{5}a - 3) \)
(iv) For \( a^4 - 3a^2 + 2 \), let's substitute \( x = a^2 \). The expression becomes \( x^2 - 3x + 2 \).
We need two numbers that multiply to 2 and add up to -3. These numbers are -1 and -2.
\( x^2 - 3x + 2 = x^2 - x - 2x + 2 \)
\( = x(x - 1) - 2(x - 1) \)
\( = (x - 1)(x - 2) \)
Now, substitute back \( x = a^2 \):
\( = (a^2 - 1)(a^2 - 2) \)
The term \( (a^2 - 1) \) can be further factored using the difference of squares formula \( (A^2 - B^2) = (A - B)(A + B) \).
\( a^2 - 1 = (a - 1)(a + 1) \).
So, the fully factored form is \( (a + 1)(a - 1)(a^2 - 2) \).
(v) For \( 8m^3 - 2m^2n - 15mn^2 \), we can first take out the common factor \( m \).
\( = m(8m^2 - 2mn - 15n^2) \)
Now, factor the quadratic expression inside the parenthesis \( (8m^2 - 2mn - 15n^2) \). Multiply the coefficients of \( m^2 \) and \( n^2 \): \( 8 \times (-15) = -120 \). We need two numbers that multiply to -120 and add up to -2. These numbers are -12 and 10.
\( = m[8m^2 - 12mn + 10mn - 15n^2] \)
\( = m[4m(2m - 3n) + 5n(2m - 3n)] \)
\( = m(2m - 3n)(4m + 5n) \)
(vi) For \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{2}{xy} \), we can rewrite this expression to recognize a standard algebraic identity.
We know the identity \( (A + B)^2 = A^2 + B^2 + 2AB \).
Here, if we let \( A = \frac{1}{x} \) and \( B = \frac{1}{y} \), then:
\( A^2 = (\frac{1}{x})^2 = \frac{1}{x^2} \)
\( B^2 = (\frac{1}{y})^2 = \frac{1}{y^2} \)
\( 2AB = 2(\frac{1}{x})(\frac{1}{y}) = \frac{2}{xy} \)
So, the given expression is exactly in the form \( A^2 + B^2 + 2AB \).
Therefore, \( \frac{1}{x^2} + \frac{1}{y^2} + \frac{2}{xy} = (\frac{1}{x} + \frac{1}{y})^2 \).
In simple words: When expressions seem complicated, first check for common factors or use substitution to make them simpler. For questions like (iv), remember to check if the factored terms can be broken down further, especially using difference of squares. Also, some expressions like (vi) might fit common algebraic formulas, so try to identify those patterns.

🎯 Exam Tip: Always look for common factors first in polynomial expressions. Be aware of common algebraic identities like perfect square trinomials and difference of squares, as they can simplify factoring significantly. When using substitution, always substitute the original variables back at the end.