Use the remainder theorem to factorise the following expression 2x3 + 9x2 + 7x - 6

Let P(x) = `2x^3 + 3x^2 - 9x - 10`

P(2) = 16 + 12 - 18 - 10

P(2) = 0

So (x - 2) is a factor

Let us divide P(x) with (x-2), we get

`(x- 2)(2x^2 + 7x + 5)`

This can be further factored to

`(x-2)(2x^2 + 5x + 2x + 5)` ……… (Split 7x into two terms, whose sum is 7x and product is10𝑥2)

`(x - 2) (2x^2 + 5x + 2x + 5)`

`(x - 2))(x(2x+5)+1(2x+5))`

(x - 2)(2x + 5)(x + 1)

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Yvette F.

Algebra

8 months, 3 weeks ago

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Use the Remainder Theorem to find the remainder when the function f(x)=x3+6x2-3x is divided by (x+5)

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Video Transcript

In this problem, we're given the coefficients of X and X and X and X and X and X and X and X and X and X and X and X and X and X and X and X and X and X and X. We're going to divide this by 5. The remainder Theorem is used. If you put five into the minus five, the result will be what the remainder is, which means I can use the synthetic division to do it. I need my coefficients with this. I have a coefficients of one, x squared is six and X is -3 and the constant is zero Let's bring down the one time -5 is -5. One -5 times 1 -5 is 6 -5. I get -8 -5 times -8 so I get 40 for the answer if I had those together. There's our results from the remainder Theorem. X squared Plus X -8 plus 40 over X plus five is the final result. There is a result

Step by step solution :

Step  1  :

Equation at the end of step  1  :

  (((2 • (x3)) +  32x2) +  7x) -  6  = 0 
 Step  2  :

Equation at the end of step  2  :

  ((2x3 +  32x2) +  7x) -  6  = 0 

Step  3  :

Checking for a perfect cube :

 3.1    2x3+9x2+7x-6  is not a perfect cube

Trying to factor by pulling out :

 3.2      Factoring:  2x3+9x2+7x-6

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1:  7x-6 
Group 2:  2x3+9x2

Pull out from each group separately :

Group 1:   (7x-6) • (1)
Group 2:   (2x+9) • (x2)

Bad news !! Factoring by pulling out fails : The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

 3.3    Find roots (zeroes) of :       F(x) = 2x3+9x2+7x-6
Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  2  and the Trailing Constant is  -6. The factor(s) are:

of the Leading Coefficient :  1,2
 of the Trailing Constant :  1 ,2 ,3 ,6 Let us test ....

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that
   2x3+9x2+7x-6 
can be divided by 3 different polynomials,including by  2x-1 

Polynomial Long Division :

 3.4    Polynomial Long Division
Dividing :  2x3+9x2+7x-6 
                              ("Dividend")
By         :    2x-1    ("Divisor")

Quotient :  x2+5x+6  Remainder:  0 

Trying to factor by splitting the middle term

 3.5     Factoring  x2+5x+6

The first term is,  x2  its coefficient is  1 .
The middle term is,  +5x  its coefficient is  5 .
The last term, "the constant", is  +6 

Step-1 : Multiply the coefficient of the first term by the constant   1 • 6 = 6

Step-2 : Find two factors of  6  whose sum equals the coefficient of the middle term, which is   5 .

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  2  and  3 
                     x2 + 2x + 3x + 6

Step-4 : Add up the first 2 terms, pulling out like factors :
                    x • (x+2)
              Add up the last 2 terms, pulling out common factors :
                    3 • (x+2)
Step-5 : Add up the four terms of step 4 :
                    (x+3)  •  (x+2)
             Which is the desired factorization

Equation at the end of step  3  :

  (x + 3) • (x + 2) • (2x - 1)  = 0 

Step  4  :

Theory - Roots of a product :

 4.1    A product of several terms equals zero. When a product of two or more terms equals zero, then at least one of the terms must be zero. We shall now solve each term = 0 separately In other words, we are going to solve as many equations as there are terms in the product Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

 4.2      Solve  :    x+3 = 0 Subtract  3  from both sides of the equation : 
                      x = -3

Solving a Single Variable Equation :

 4.3      Solve  :    x+2 = 0 Subtract  2  from both sides of the equation : 
                      x = -2

Solving a Single Variable Equation :

 4.4      Solve  :    2x-1 = 0 Add  1  to both sides of the equation : 
                      2x = 1
Divide both sides of the equation by 2:
                     x = 1/2 = 0.500

Supplement : Solving Quadratic Equation Directly

Solving    x2+5x+6  = 0   directly 

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

 5.1      Find the Vertex of   y = x2+5x+6Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 1 , is positive (greater than zero). Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is  -2.5000  Plugging into the parabola formula  -2.5000  for  x  we can calculate the  y -coordinate : 
  y = 1.0 * -2.50 * -2.50 + 5.0 * -2.50 + 6.0
or   y = -0.250

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2+5x+6
Axis of Symmetry (dashed)  {x}={-2.50} 
Vertex at  {x,y} = {-2.50,-0.25} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-3.00, 0.00} 
Root 2 at  {x,y} = {-2.00, 0.00} 

Solve Quadratic Equation by Completing The Square

 5.2     Solving   x2+5x+6 = 0 by Completing The Square .Subtract  6  from both side of the equation :
   x2+5x = -6

Now the clever bit: Take the coefficient of  x , which is  5 , divide by two, giving  5/2 , and finally square it giving  25/4

Add  25/4  to both sides of the equation :
  On the right hand side we have :
   -6  +  25/4    or,  (-6/1)+(25/4) 
  The common denominator of the two fractions is  4   Adding  (-24/4)+(25/4)  gives  1/4 
  So adding to both sides we finally get :
   x2+5x+(25/4) = 1/4

Adding  25/4  has completed the left hand side into a perfect square :
   x2+5x+(25/4)  =
   (x+(5/2)) • (x+(5/2))  =
  (x+(5/2))2
Things which are equal to the same thing are also equal to one another. Since
   x2+5x+(25/4) = 1/4 and
   x2+5x+(25/4) = (x+(5/2))2
then, according to the law of transitivity,
   (x+(5/2))2 = 1/4

We'll refer to this Equation as  Eq. #5.2.1  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x+(5/2))2  is
   (x+(5/2))2/2 =
  (x+(5/2))1 =
   x+(5/2)

Now, applying the Square Root Principle to  Eq. #5.2.1  we get:
   x+(5/2) = √ 1/4

Subtract  5/2  from both sides to obtain:
   x = -5/2 + √ 1/4

Since a square root has two values, one positive and the other negative
   x2 + 5x + 6 = 0
   has two solutions:
  x = -5/2 + √ 1/4
   or
  x = -5/2 - √ 1/4

Note that  √ 1/4 can be written as
  √ 1  / √ 4   which is 1 / 2

Solve Quadratic Equation using the Quadratic Formula

 5.3     Solving    x2+5x+6 = 0 by the Quadratic Formula .According to the Quadratic Formula,  x  , the solution for   Ax2+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B2-4AC
  x =   ————————
                      2A   In our case,  A   =     1
                      B   =    5
                      C   =   6 Accordingly,  B2  -  4AC   =
                     25 - 24 =
                     1Applying the quadratic formula :

               -5 ± √ 1
   x  =    —————
                    2So now we are looking at:
           x  =  ( -5 ± 1) / 2

Two real solutions:

x =(-5+√1)/2=-2.000

or:

x =(-5-√1)/2=-3.000

Three solutions were found :

1.  x = 1/2 = 0.500
   
2.  x = -2
3.  x = -3