Answer :
We start with the polynomial
[tex]$$
2x^4 - 24x^2 + 70.
$$[/tex]
Step 1. Factor out the greatest common factor
Notice that all terms are even, so we can factor out a [tex]$2$[/tex]:
[tex]$$
2x^4 - 24x^2 + 70 = 2\left(x^4 - 12x^2 + 35\right).
$$[/tex]
Step 2. Recognize the quadratic form in [tex]$x^2$[/tex]
The expression inside the parentheses, [tex]$x^4 - 12x^2 + 35$[/tex], is a quadratic in [tex]$x^2$[/tex]. To see this clearly, let
[tex]$$
y = x^2.
$$[/tex]
Then
[tex]$$
x^4 - 12x^2 + 35 = y^2 - 12y + 35.
$$[/tex]
Step 3. Factor the quadratic
We now factor the quadratic
[tex]$$
y^2 - 12y + 35.
$$[/tex]
We are looking for two numbers that multiply to [tex]$35$[/tex] and add to [tex]$-12$[/tex]. These numbers are [tex]$-5$[/tex] and [tex]$-7$[/tex], because
[tex]$$
(-5)(-7) = 35 \quad \text{and} \quad (-5) + (-7) = -12.
$$[/tex]
Thus, the quadratic factors as
[tex]$$
y^2 - 12y + 35 = (y - 5)(y - 7).
$$[/tex]
Step 4. Substitute back [tex]$x^2$[/tex] for [tex]$y$[/tex]
Replacing [tex]$y$[/tex] by [tex]$x^2$[/tex], we obtain
[tex]$$
x^4 - 12x^2 + 35 = (x^2 - 5)(x^2 - 7).
$$[/tex]
Step 5. Write the final factorization
Substituting this back into our original expression, we have
[tex]$$
2x^4 - 24x^2 + 70 = 2(x^2 - 5)(x^2 - 7).
$$[/tex]
Thus, the complete factorization of the trinomial is:
[tex]$$
\boxed{2(x^2 - 5)(x^2 - 7)}.
$$[/tex]
[tex]$$
2x^4 - 24x^2 + 70.
$$[/tex]
Step 1. Factor out the greatest common factor
Notice that all terms are even, so we can factor out a [tex]$2$[/tex]:
[tex]$$
2x^4 - 24x^2 + 70 = 2\left(x^4 - 12x^2 + 35\right).
$$[/tex]
Step 2. Recognize the quadratic form in [tex]$x^2$[/tex]
The expression inside the parentheses, [tex]$x^4 - 12x^2 + 35$[/tex], is a quadratic in [tex]$x^2$[/tex]. To see this clearly, let
[tex]$$
y = x^2.
$$[/tex]
Then
[tex]$$
x^4 - 12x^2 + 35 = y^2 - 12y + 35.
$$[/tex]
Step 3. Factor the quadratic
We now factor the quadratic
[tex]$$
y^2 - 12y + 35.
$$[/tex]
We are looking for two numbers that multiply to [tex]$35$[/tex] and add to [tex]$-12$[/tex]. These numbers are [tex]$-5$[/tex] and [tex]$-7$[/tex], because
[tex]$$
(-5)(-7) = 35 \quad \text{and} \quad (-5) + (-7) = -12.
$$[/tex]
Thus, the quadratic factors as
[tex]$$
y^2 - 12y + 35 = (y - 5)(y - 7).
$$[/tex]
Step 4. Substitute back [tex]$x^2$[/tex] for [tex]$y$[/tex]
Replacing [tex]$y$[/tex] by [tex]$x^2$[/tex], we obtain
[tex]$$
x^4 - 12x^2 + 35 = (x^2 - 5)(x^2 - 7).
$$[/tex]
Step 5. Write the final factorization
Substituting this back into our original expression, we have
[tex]$$
2x^4 - 24x^2 + 70 = 2(x^2 - 5)(x^2 - 7).
$$[/tex]
Thus, the complete factorization of the trinomial is:
[tex]$$
\boxed{2(x^2 - 5)(x^2 - 7)}.
$$[/tex]
