Question 985473


a.  To determine whether  x-5  is a factor of  P(x),  calculate  P(5),  i.e.  simply substitute the value of  5  into the polynomial.  You will get 


P(5) = {{{5^3 - 2*5^2 - 13*5 - 10}}} = 125 - 50 - 65 - 10 = 0.


According wi the &nbsp;<B>Remainder Theorem</B>&nbsp; (see, &nbsp;for example, &nbsp;the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Divisibility-of-polynomial-f%28x%29-by-binomial-x-a.lesson>Divisibility of polynomial f(x) by binomial x-a</A>&nbsp; in this site), &nbsp;the binom &nbsp;x-5&nbsp; is a factor of the polynomial &nbsp;P(x).



b. &nbsp;Make the long division of the polynomial &nbsp;P(x) = {{{x^3-2x^2-13x-10}}}&nbsp; by &nbsp;{{{(x-5)}}}. &nbsp;You will get 


{{{x^3-2x^2-13x-10}}} = {{{(x-5)}}}.{{{(x^2 + 3x + 2)}}}. 


So, &nbsp;the polynomial &nbsp;{{{(x^2 + 3x + 2)}}}&nbsp; is another factor of the polynomial &nbsp;P(x). 



c. &nbsp;The quadratic polynomial &nbsp;{{{x^2 + 3x + 2}}}&nbsp; has the roots &nbsp;{{{x[1]}}} = -1&nbsp; and &nbsp;{{{x[2]}}} = -2&nbsp; (use the quadratic formula &nbsp;<A HREF=http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A>&nbsp; or the Vieta's theorem &nbsp;<A HREF=http://www.algebra.com/algebra/homework/quadratic/lessons/Solving-quadratic-equations-without-quadratic-formula.lesson>Solving quadratic equations without quadratic formula</A>, &nbsp;lessons in this site)


It implies that &nbsp;{{{x^2 + 3x + 2}}} = {{{(x+1)}}}.{{{(x+2)}}}. 


Therefore, 


P(x) = {{{(x-5)}}}.{{{(x+1)}}}.{{{(x+2)}}} 


is the complete factorization of the polynomial &nbsp;P(x) = {{{x^3-2x^2-13x-10}}}.



d. &nbsp;Since P(x) = {{{x^3-2x^2-13x-10}}} = {{{(x-5)}}}.{{{(x+1)}}}.{{{(x+2)}}}, &nbsp;the roots of the polynomial {{{x^3-2x^2-13x-10}}} are 5, -1 and -2. 



The solution is completed.