Example Question - factorization of trinomials
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Factoring Trinomials
The given trinomial is \(3x^2 + 17x + 10\). To factor it we need to find two numbers that multiply to \(3x^2 \cdot 10\) (the coefficient of \(x^2\) times the constant term) and add up to \(17x\) (the coefficient of the middle term). The two numbers that fit this criterion are 2 and 15, since: \(2 \cdot 15 = 30\), and \(30\) is the product of the coefficient of \(x^2\) (3) and the constant term (10). \(2 + 15 = 17\), which is the coefficient of the middle term. Now we need to split the middle term, \(17x\), into two terms using the numbers 2 and 15: \(3x^2 + 2x + 15x + 10\) Next, we group terms: \((3x^2 + 2x) + (15x + 10)\) Factor out the common factor from each group: \(x(3x + 2) + 5(3x + 2)\) We now have a common binomial factor of \((3x + 2)\) which we can factor out: \((3x + 2)(x + 5)\) Therefore, the trinomial \(3x^2 + 17x + 10\) factors to \((3x + 2)(x + 5)\).
Factoring Trinomials
The question is asking to factor the trinomial \(2x^2 + 11x + 14\). To factor it, one of the methods is to look for two numbers that both add up to the coefficient of the \(x\) term (which is 11 in this case) and multiply to the product of the coefficient of \(x^2\) term and the constant term (which is \(2 \times 14 = 28\)). So, we need two numbers that add up to 11 and multiply to 28. These two numbers are 4 and 7, because: \[4 + 7 = 11\] \[4 \times 7 = 28\] Now we can rewrite the middle term (11x) using 4 and 7: \[2x^2 + 4x + 7x + 14\] Next, let's factor by grouping: \[2x(x + 2) + 7(x + 2)\] Now, we can take out the common factor \((x + 2)\): \[(2x + 7)(x + 2)\] Therefore, the factored form of \(2x^2 + 11x + 14\) is \((2x + 7)(x + 2)\).
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