We expand (2x - a)(bx + 2) = 2bx² + (4 - ab)x - 2a and compare coefficients with 10x² - 11x - 6. We get 2b = 10 ⇒ b = 5, and -2a = -6 ⇒ a = 3, which also satisfies 4 - (3)(5) = -11.

To factor 6x² + 7x + 2, we find two numbers that multiply to 6 × 2 = 12 and add up to 7. These numbers are 4 and 3. Splitting the middle term gives 2x(3x + 2) + 1(3x + 2) = (3x + 2)(2x + 1).

EXERCISE SET 4.4 1. Fill in the blanks to complete the following identities: (i) $s^2 - 11s + 24 = (\underline{\hspace{1.5cm}})(\underline{\hspace{1.5cm}})$ (ii) $(\underline{\hspace{1.5cm}})(x + 1) = (3x^2 - 4x - 7)$ (iii) $10x^2 - 11x - 6 = (2x - \underline{\hspace{1cm}})(\underline{\hspace{1cm}} + 2)$ (iv) $6x^2 + 7x + 2 = (\underline{\hspace{1.5cm}})(\underline{\hspace{1.5cm}})$ 2. Select and use the identity that will help you...

Answer: (i) $(s - 8)(s - 3)$ (ii) $(3x - 7)$ (iii) $a = 3$ and $b = 5x$ , giving $(2x - 3)(5x + 2)$ (iv) $(3x + 2)(2x + 1)$

Step-by-step solution

Step 1: Factorize the quadratic expression in part (i)

To factor $s^2 - 11s + 24$ , we search for two integers whose product is $24$ and whose sum is $-11$ . The numbers are $-8$ and $-3$ , yielding the factorization $(s - 8)(s - 3)$ .

Step 2: Find the missing factor in part (ii)

We can factor $3x^2 - 4x - 7$ by splitting the middle term into $-7x + 3x$ . Grouping terms gives $x(3x - 7) + 1(3x - 7) = (3x - 7)(x + 1)$ , so the missing term is $3x - 7$ .

Step 3: Complete the blanks in part (iii)

We expand $(2x - a)(bx + 2) = 2bx^2 + (4 - ab)x - 2a$ and compare coefficients with $10x^2 - 11x - 6$ . We get $2b = 10 \implies b = 5$ , and $-2a = -6 \implies a = 3$ , which also satisfies $4 - (3)(5) = -11$ .

Step 4: Factorize the quadratic expression in part (iv)

To factor $6x^2 + 7x + 2$ , we find two numbers that multiply to $6 \times 2 = 12$ and add up to $7$ . These numbers are $4$ and $3$ . Splitting the middle term gives $2x(3x + 2) + 1(3x + 2) = (3x + 2)(2x + 1)$ .

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