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13 December, 09:14

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

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  1. 13 December, 12:50
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    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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