Question 985625


Factor the polynomial  P(x).  Then solve the equation  P(x)=0.


1.  P(x) = {{{x^3+4x^2+x-6}}}
2.  P(x) = {{{x^3-6x^2-x+6}}}
3.  P(x) = {{{x^3-x^2-x+1}}}
4.  P(x) = {{{2x^3-3x^2-3x+2}}}
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1.  P(x) = {{{x^3 + 4x^2 + x - 6}}}.


   The integer  1  is the root.  Indeed,  P(1) = {{{1^3 + 4*1^2 + 1 - 6}}} = {{{1 + 4 + 1 - 6}}} = {{{0}}}.

It means that the binomial &nbsp;(x-1)&nbsp; divides the polynomial &nbsp;P(x): &nbsp;P(x) = (x-1)*Q(x), &nbsp;where &nbsp;Q(x)&nbsp; is a quadratic polynomial. &nbsp;(See the &nbsp;<B>Remainder Theorem</B>&nbsp; in the lesson 

<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). 


If you make the long division, &nbsp;you will get &nbsp;Q(x) = {{{x^2 + 5x + 6}}}. 


The quadratic polynomial &nbsp;Q(x)&nbsp; has the roots &nbsp;-2&nbsp; and &nbsp;-3: &nbsp;Q(-2) = 0 &nbsp;and &nbsp;Q(-3) = 0.

You can use the quadratic formula to find the roots &nbsp;(see the lesson &nbsp;<A HREF =http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson>Introduction into Quadratic Equations</A>&nbsp; in this site)&nbsp; or 

the &nbsp;<B>Vieta's Theorem</B>&nbsp; (see the lesson &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;You also can  check it directly. 


This means that the binomials &nbsp;(x+2)&nbsp; and &nbsp;(x+3)&nbsp; divide the polynomial &nbsp;Q(x), &nbsp;so &nbsp;Q(x) = (x+2)*(x+3). 

You also can check this factorization immediately. 


It implies that the polynomial &nbsp;P(x)&nbsp; has the factorization &nbsp;P(x) = (x-1)*(x+2)*(x+3). 

Hence, &nbsp;its roots are &nbsp;1, &nbsp;-2 and &nbsp;-3.



2. &nbsp;P(x) = {{{x^3 - 6x^2 - x + 6}}}. 


Re-group: &nbsp;P(x) = {{{(x^3 - 6x^2)}}} - {{{(x - 6)}}} = {{{x^2*(x-6) - (x-6)}}} = {{{(x-6)*(x^2-1)}}} = (x-6)*(x-1)*(x+1).


The roots of the polynomial &nbsp;P(x)&nbsp; are &nbsp;6, &nbsp;1&nbsp; and &nbsp;-1. 



3. &nbsp;P(x) = {{{x^3 - x^2 - x + 1}}}.  


Re-group: &nbsp;P(x) = {{{(x^3 - x^2)}}} - {{{(x - 1)}}} = {{{x^2*(x-1) - (x-1)}}} = {{{(x-1)*(x^2-1)}}} = (x-1)*(x-1)*(x+1).


The roots of the polynomial &nbsp;P(x)&nbsp; are &nbsp;1 &nbsp;(multiplicity 2), &nbsp;and &nbsp;-1. 



4. P(x) = {{{2x^3 - 3x^2 - 3x + 2}}}. 


One root is &nbsp;2: &nbsp;P(2) = 0 &nbsp;(check it yourself). 


Hence, &nbsp;P(x)&nbsp; is divided by &nbsp;(x-2): &nbsp;P(x) = (x-2)*Q(x), &nbsp;where &nbsp;Q(x)&nbsp; is a quadratic polynomial. &nbsp;(By the same reason as in the &nbsp;n.1&nbsp; above). 


Long division gives &nbsp;Q(x) = {{{2x^2 + x - 1}}}. 


Hence, &nbsp;P(x) = {{{(x-2)*(2x^2 + x - 1)}}}. 


The quadratic polynomial &nbsp;{{{2x^2 + x - 1}}}&nbsp; has the roots &nbsp;-1 &nbsp;and &nbsp;{{{1/2}}}. &nbsp;You can find them using the same methods as in the &nbsp;n.1&nbsp; above.

So, &nbsp;Q(x) = {{{2x^2 + x - 1}}} = {{{2*(x+1)*(x-1/2)}}}. 


Thus &nbsp;P(x) = {{{2*(x-2)*(x+1)*(x-1/2)}}} &nbsp;is the final factorization of the polynomial &nbsp;P(x)&nbsp; over the real domain. 


The polynomial &nbsp;P(x)&nbsp; has the roots &nbsp;2, &nbsp;-1 &nbsp;and &nbsp;{{{1/2}}}.