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How to factorise using the power of a?
Date Posted: 3 years ago
Views: 300
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Date Posted:
3 years ago
Method 1 (not using power of a, only using (a + b)² = x² + 2ab + b² and (a + b)(a - b) = a² - b²)
Rerranging, x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= x³ + 2ax² + x² + a²x + 2ax - x + a² - 1
= x³ + 2ax² + a²x - x + x² + 2ax + a² - 1
= x(x² + 2ax + a² - 1) + x² + 2ax + a² - 1
= (x + 1)(x² + 2ax + a² - 1)
= (x + 1)((x + a)² - 1²)
= (x + 1)(x + a + 1)(x + a - 1)
Method 2 (using power of a, long division and cross method of factorisation not required at all)
x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= a²x + a² + 2ax² + 2ax + x³ + x² - x - 1
= a²(x + 1) + 2ax(x + 1) + x²(x + 1) - (x + 1)
= (x + 1)(a² + 2ax + x² - 1)
= (x + 1)((a + x)² - 1²)
= (x + 1)(a + x + 1)(a + x - 1)
Rerranging, x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= x³ + 2ax² + x² + a²x + 2ax - x + a² - 1
= x³ + 2ax² + a²x - x + x² + 2ax + a² - 1
= x(x² + 2ax + a² - 1) + x² + 2ax + a² - 1
= (x + 1)(x² + 2ax + a² - 1)
= (x + 1)((x + a)² - 1²)
= (x + 1)(x + a + 1)(x + a - 1)
Method 2 (using power of a, long division and cross method of factorisation not required at all)
x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= a²x + a² + 2ax² + 2ax + x³ + x² - x - 1
= a²(x + 1) + 2ax(x + 1) + x²(x + 1) - (x + 1)
= (x + 1)(a² + 2ax + x² - 1)
= (x + 1)((a + x)² - 1²)
= (x + 1)(a + x + 1)(a + x - 1)
Method 3 (variant of Method 1)
x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= x³ + x² + 2ax² + 2ax + (a² - 1)x + a² - 1
= x²(x + 1) + 2ax(x + 1) + (a² - 1)(x + 1)
= (x² + 2ax + a² - 1)(x + 1)
= (x + 1)((x + a)² - 1²)
= (x + 1)(x + a + 1)(x + a - 1)
x³ + (2a + 1)x² + (a² + 2a - 1)x + (a² - 1)
= x³ + x² + 2ax² + 2ax + (a² - 1)x + a² - 1
= x²(x + 1) + 2ax(x + 1) + (a² - 1)(x + 1)
= (x² + 2ax + a² - 1)(x + 1)
= (x + 1)((x + a)² - 1²)
= (x + 1)(x + a + 1)(x + a - 1)
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