Chapter 5: Problem 16

Factorise (a) \(2 x^{2}-11 x+5\) (b) \(2 x^{2}+13 x+15\)

Short Answer

Expert verified

(a) \(2x^2 - 11x + 5\) (b) \(2x^2 + 13x + 15\) Answer: (a) \((x - 5)(2x - 1)\) (b) \((x + 5)(2x + 3)\)

Step by step solution

(a) Factorise \(2x^2 - 11x + 5\)

1. Identify the coefficients: \(a=2, b=-11, c=5\) 2. We need to find two numbers that multiply to give \((2)(5) = 10\) and add up to \(-11\). These numbers are \(-1\) and \(-10\). 3. Rewrite the middle term using the two numbers we found: \(2x^2 - 11x + 5 = 2x^2 - x - 10x + 5\) 4. Factor by grouping: \((2x^2 - x) - (10x - 5) = x(2x - 1) - 5(2x - 1) = (x - 5)(2x - 1)\) So, the factorised form of \((a)\) is \((x - 5)(2x - 1)\).

(b) Factorise \(2x^2 + 13x + 15\)

1. Identify the coefficients: \(a=2, b=13, c=15\) 2. We need to find two numbers that multiply to give \((2)(15) = 30\) and add up to \(13\). These numbers are \(3\) and \(10\). 3. Rewrite the middle term using the two numbers we found: \(2x^2 + 13x + 15 = 2x^2 + 3x + 10x + 15\) 4. Factor by grouping: \((2x^2 + 3x) + (10x + 15) = x(2x + 3) + 5(2x + 3) = (x + 5)(2x + 3)\) So, the factorised form of \((b)\) is \((x + 5)(2x + 3)\).

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Factorise (a) \(2 x^{2}-11 x+5\) (b) \(2 x^{2}+13 x+15\)

Short Answer

Step by step solution

(a) Factorise \(2x^2 - 11x + 5\)

(b) Factorise \(2x^2 + 13x + 15\)

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