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Chapter 0: Problem 68

Factor completely, or state that the polynomial is prime. $$6 x^{2}-18 x-60$$

Short Answer

Expert verified
The completely factored form of the polynomial \(6 x^{2}-18 x-60\) is \(6(x-5)(x+2)\)

Step by step solution

01

Identify the Greatest Common Factor (GCF)

Looking at each term of the polynomial \(6 x^{2}-18 x-60\), you can see that all coefficients (6,-18, and -60) can be evenly divided by 6. Therefore, the GCF is 6.
02

Factor out the GCF

Divide each term in the polynomial by the GCF and write the polynomial as the product of the GCF and the resulting expression. Hence, factoring out 6 from \(6 x^{2}-18 x-60\) gives \(6(x^{2}-3 x-10)\)
03

Factor the resulting polynomial

The resulting polynomial \(x^{2}-3 x-10\) is a quadratic polynomial, which is factorable. The quadratic can be factored by finding two numbers that multiply to -10 (the last term) and add to -3 (the coefficient of the middle term). The numbers that satisfy this are -5 and 2. Hence, the factored form of \(x^{2}-3 x-10\) is \((x-5)(x+2)\)
04

Write down the completely factored form of the polynomial

Combine the factored expressions from Step 2 and Step 3. The completely factored form of the original polynomial \(6 x^{2}-18 x-60\) is \(6(x-5)(x+2)\)

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