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

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