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Factorise x³-x²-x+3
expand (2a-3b+6c)²​

User Anupam Roy
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2 Answers

23 votes
23 votes

To factorize x³-x²-x+3, we can use the factoring method called "guess and check." This involves trying different factor pairs that could multiply to give the constant term (3 in this case) and adding or subtracting them to give the coefficient of the x term (-1 in this case).

One possible factorization of x³-x²-x+3 is (x-1)(x²+x+3). We can verify that this factorization is correct by expanding (x-1)(x²+x+3) using the distributive property:

(x-1)(x²+x+3) = x(x²+x+3) - 1(x²+x+3)

= x³ + x² + 3x - x² - x - 3

= x³ - x² - x + 3

Therefore, x³-x²-x+3 can be factored as (x-1)(x²+x+3).

To expand (2a-3b+6c)², we can use the binomial expansion formula:

(2a-3b+6c)² = (2a)² + 2(2a)(-3b) + (-3b)² + 2(2a)(6c) + 2(-3b)(6c) + (6c)²

= 4a² - 12ab + 9b² + 12ac - 18bc + 36c²

= 4a² - 6ab + 3b² + 12ac - 6bc + 12c²

Therefore, (2a-3b+6c)² can be expanded as 4a² - 6ab + 3b² + 12ac - 6bc

User Damio
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12 votes
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factorize: 3-x-x^2+x^3
expand: 4a^2+9b^2+36c^2-12ab+24ac-36bc
User Gibson
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3.1k points