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Equations- Factorization of a polynomial equation – Exercise 5604

Exercise

Factor the polynomial equation

x^4-3x^2-4=0

Final Answer


(x-2)(x+2)(x^2+1)=0

Solution

First solution:

x^4-3x^2-4=0

To reach a quadratic equation, we define a new variable:

y=x^2

We set the new variable:

y^2-3y-4=0

We got a quadratic equation. Our equation coefficients are

a=1, b=-3, c=-4

We solve it with the quadratic formula. Putting the coefficients in the formula gives us

y_{1,2}=\frac{3\pm \sqrt{{(-3)}^2-4\cdot 1\cdot (-4)}}{2\cdot 1}=

=\frac{3\pm \sqrt{25}}{2}=

=\frac{3\pm 5}{2}

Hence, we get the solutions:

y_1=\frac{3+ 5}{2}=\frac{8}{2}=4

y_2=\frac{3- 5}{2}=\frac{-2}{2}=-1

Thus, the factorizing of the quadratic equation is

y^2-3y-4=

=(y-4)(y+1)

Going back to the original variable we get

=(x^2-4)(x^2+1)

We break down the first factor using Short Multiplication Formulas (third formula) and get the factorizing

=(x-2)(x+2)(x^2+1)

Second solution:

x^4-3x^2-4=

We express the second expression as sum and we get

=x^4-4x^2+x^2-4=

We extract a common factor from the first two expressions:

=x^2(x^2-4)+(x^2-4)=

Once again we extract a common factor – this time the expression in parentheses:

=(x^2-4)(x^2+1)=

We break down the first factor using Short Multiplication Formulas (third formula) and get the factorizing:

=(x-2)(x+2)(x^2+1)

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