 # Equations- Factorization of a polynomial equation – Exercise 5604

Exercise

Factor the polynomial equation

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

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

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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