Question

examples:
factor each polynomial expression completely

1. $x^4 + 8x^2 + 7$
2. $x^4 - x^2 - 72$
3. $5x^4 + 25x^2 - 30$
4. $2x^4 - 7x^2 + 3$
5. $9x^4 + 6x^2 - 63$
6. $32x^4 + 64x^2 - 18$

Explanation:

Let's solve problem 2: \(x^4 - x^2 - 72\)

Step1: Let \(y = x^2\)

Substitute \(y\) into the polynomial, we get \(y^2 - y - 72\)

Step2: Factor the quadratic in \(y\)

We need two numbers that multiply to \(-72\) and add to \(-1\). The numbers are \(-9\) and \(8\). So, \(y^2 - y - 72=(y - 9)(y + 8)\)

Step3: Substitute back \(y = x^2\)

We have \((x^2 - 9)(x^2 + 8)\)

Step4: Factor \(x^2 - 9\) (difference of squares)

\(x^2 - 9=(x - 3)(x + 3)\)
So the completely factored form is \((x - 3)(x + 3)(x^2 + 8)\)

Step1: Let \(y = x^2\)

Substitute \(y\) into the polynomial, we get \(2y^2 - 7y + 3\)

Step2: Factor the quadratic in \(y\)

We need two numbers that multiply to \(2\times3 = 6\) and add to \(-7\). The numbers are \(-6\) and \(-1\). Rewrite the middle term: \(2y^2-6y - y + 3\). Group and factor: \(2y(y - 3)-1(y - 3)=(2y - 1)(y - 3)\)

Step3: Substitute back \(y = x^2\)

We have \((2x^2 - 1)(x^2 - 3)\)

Step1: Factor out the GCF (3)

\(3(3x^4 + 2x^2 - 21)\)

Step2: Let \(y = x^2\), the quadratic in \(y\) is \(3y^2 + 2y - 21\)

We need two numbers that multiply to \(3\times(-21)=-63\) and add to \(2\). The numbers are \(9\) and \(-7\). Rewrite the middle term: \(3y^2+9y - 7y - 21\). Group and factor: \(3y(y + 3)-7(y + 3)=(3y - 7)(y + 3)\)

Step3: Substitute back \(y = x^2\) and factor out the GCF part

\(3(3x^2 - 7)(x^2 + 3)\)

Answer:

\((x - 3)(x + 3)(x^2 + 8)\)

Now problem 4: \(2x^4 - 7x^2 + 3\)