Using a given zero to write a polynomial as a product of linear factors

Algebra
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For the polynomial below, 3 is a zero. 

f(x)= x^3 - x^2 - 12x + 18

Express f(x) as a product of linear factors.

Oct 16th, 2015

Thank you for the opportunity to help you with your question!

 by factoring, we are looking for "zeros" of polynomial, i.e. values of x, for which the polynomial is equal to zero. Underneath the hood, these values are points on x-axis where a graph of polynomial crosses it, hence, the value of y in them is 0. 

A linear factor has a form of (x-a). When we set this factor equal to zero, we get 

x - a = 0 
x = a 

Hence, for some value a, the polynomial is equal to zero. ( f(a) = 0 ). We are already told that the first such value is 3. To check, you can plug in 3 and then solve, to get 0. 
So our first factor is (x - 3). Now, we have to find the quadratic factor, and then, factorize it into linear factors. 

(x - 3)(ax^2 + bx + c) = 0 

When we multiple these two factors, we must get the original polynomial. First, we multiply x by ax^2, which gives ax^3. In the original polynomial, the coefficient of x^3 is 1, so a = 1. 

(x - 3)(x^2 + bx + c) = 0 

Now, we have to deal with b. When we multiply x with bx, we get bx^2. However, we also get some x^2 by multiplying (-3) with (x^2). Hence, we can form an equation, such as -3x^2 + bx^2 = 3x^2 (since the coefficient of x^2 in the original equation is 3). By solving, we see that b = 6. 

(x - 3)(x^2 + 6x + c) = 0 

Again, we get x terms by multiplying x with c and (-3) with 6x. Again, we can form an equation, cx - 18x = -12x. By solving, we get c = 6. (Alternatively, we can look at constant term, which is -18, set an equation such as (-3) * c = -18, and we again see that c = 6. However, not all polynomials have constant term, so it is better to factorize using x terms) 

(x - 3)(x^2 + 6x + 6) = 0. 

Now we have to factor quadratic. We would normally look for such values of a and b, that a*b = 6 and a+b = 6 as well, and form two linear factors (x - a)(x - b). However, such values are not integers, so we are going to use formula x = (-b +- √(b^2 - 4ac)) / 2a. 

x = (-6 +- √36 - 24) / 2 
x = (-6 +-3.67) / 2 
x = -1.165 and -4.835 

(x - 3)(x - 1.165)(x - 4.835) 

It is unusual to factor quadratic using rational numbers, but whatever. This method is called "by inspection", it may seem overwhelming, but after some practice you are able to do this whole thing inside your head. 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 16th, 2015

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