Factor the folowing equation completely and be sure to show all work: x⁴-10x²+4.

x4-10x2+9=0  

Four solutions were found :

x = 3

x = -3

x = 1

x = -1

Step by step solution :

Step  1  :

Equation at the end of step  1  :

 ((x4) -  (2•5x2)) +  9  = 0  

Step  2  :

Trying to factor by splitting the middle term

2.1     Factoring  x4-10x2+9  

The first term is,  x4  its coefficient is  1 .

The middle term is,  -10x2  its coefficient is  -10 .

The last term, "the constant", is  +9  

Step-1 : Multiply the coefficient of the first term by the constant   1 • 9 = 9  

Step-2 : Find two factors of  9  whose sum equals the coefficient of the middle term, which is   -10 .

     -9    +    -1    =    -10    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  -1  

                    x4 - 9x2 - 1x2 - 9

Step-4 : Add up the first 2 terms, pulling out like factors :

                   x2 • (x2-9)

             Add up the last 2 terms, pulling out common factors :

                    1 • (x2-9)

Step-5 : Add up the four terms of step 4 :

                   (x2-1)  •  (x2-9)

            Which is the desired factorization

Trying to factor as a Difference of Squares :

2.2      Factoring:  x2-1  

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1

Check :  x2  is the square of  x1  

Factorization is :       (x + 1)  •  (x - 1)  

Trying to factor as a Difference of Squares :

2.3      Factoring:  x2 - 9  

Check : 9 is the square of 3

Check :  x2  is the square of  x1  

Factorization is :       (x + 3)  •  (x - 3)  

Equation at the end of step  2  :

 (x + 1) • (x - 1) • (x + 3) • (x - 3)  = 0  

Step  3  :

Theory - Roots of a product :

3.1    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 :

3.2      Solve  :    x+1 = 0  

Subtract  1  from both sides of the equation :  

                     x = -1

Solving a Single Variable Equation :

3.3      Solve  :    x-1 = 0  

Add  1  to both sides of the equation :  

                     x = 1

Solving a Single Variable Equation :

3.4      Solve  :    x+3 = 0  

Subtract  3  from both sides of the equation :  

                     x = -3

Solving a Single Variable Equation :

3.5      Solve  :    x-3 = 0  

Add  3  to both sides of the equation :  

                     x = 3

Supplement : Solving Quadratic Equation Directly

Solving    x4-10x2+9  = 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

Solving a Single Variable Equation :

Equations which are reducible to quadratic :

4.1     Solve   x4-10x2+9 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x2  transforms the equation into :

w2-10w+9 = 0

Solving this new equation using the quadratic formula we get two real solutions :

  9.0000  or   1.0000

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of

  x4-10x2+9 = 0

 are either :  

 x =√ 9.000 = 3.00000  or :

 x =√ 9.000 = -3.00000  or :

 x =√ 1.000 = 1.00000  or :

 x =√ 1.000 = -1.00000

Four solutions were found :

x = 3

x = -3

x = 1

x = -1

Processing ends successfully

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πr^2/2

where r is the radius of the pizza.

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