Answer :
To factor and find the zeros of the polynomial function [tex]\( f(x) = 5x^4 - 34x^2 + 45 \)[/tex], we can follow these steps:
### Step 1: Factoring the Polynomial
First, we need to factor the given polynomial expression.
The polynomial [tex]\( 5x^4 - 34x^2 + 45 \)[/tex] can be thought of in terms of a quadratic form by letting [tex]\( u = x^2 \)[/tex]. Thus, we rewrite the expression as:
[tex]\[ 5u^2 - 34u + 45 \][/tex]
Next, we can factor this quadratic polynomial. We need to find factors such that the product equals the constant term (45) and the middle term (-34) is split appropriately.
[tex]\[ 5u^2 - 34u + 45 = 0 \][/tex]
Following the factorization process, we find:
[tex]\[ 5u^2 - 34u + 45 = (u - 5)(5u - 9) \][/tex]
Since we substituted [tex]\( u = x^2 \)[/tex], substituting back gives:
[tex]\[ 5x^4 - 34x^2 + 45 = (x^2 - 5)(5x^2 - 9) \][/tex]
### Step 2: Solving for Zeros
The zeros of the polynomial function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex]. Therefore, we set each factor equal to zero and solve for [tex]\( x \)[/tex].
1. [tex]\( x^2 - 5 = 0 \)[/tex]
[tex]\[
x^2 = 5 \quad \implies \quad x = \pm\sqrt{5}
\][/tex]
2. [tex]\( 5x^2 - 9 = 0 \)[/tex]
[tex]\[
5x^2 = 9 \quad \implies \quad x^2 = \frac{9}{5} \quad \implies \quad x = \pm\sqrt{\frac{9}{5}} \quad \implies \quad x = \pm\frac{3}{\sqrt{5}} = \pm\frac{3\sqrt{5}}{5}
\][/tex]
### Step 3: Listing All Zeros
Combining the solutions from the steps above, the zeros of the polynomial [tex]\( f(x) = 5x^4 - 34x^2 + 45 \)[/tex] are:
[tex]\[
x = -\sqrt{5}, \quad x = \sqrt{5}, \quad x = -\frac{3\sqrt{5}}{5}, \quad x = \frac{3\sqrt{5}}{5}
\][/tex]
### Summary
The factorized form of the polynomial [tex]\( f(x) = 5x^4 - 34x^2 + 45 \)[/tex] is:
[tex]\[ f(x) = (x^2 - 5)(5x^2 - 9) \][/tex]
The zeros of the polynomial are:
[tex]\[ x = -\sqrt{5}, \quad x = \sqrt{5}, \quad x = -\frac{3\sqrt{5}}{5}, \quad x = \frac{3\sqrt{5}}{5} \][/tex]
### Step 1: Factoring the Polynomial
First, we need to factor the given polynomial expression.
The polynomial [tex]\( 5x^4 - 34x^2 + 45 \)[/tex] can be thought of in terms of a quadratic form by letting [tex]\( u = x^2 \)[/tex]. Thus, we rewrite the expression as:
[tex]\[ 5u^2 - 34u + 45 \][/tex]
Next, we can factor this quadratic polynomial. We need to find factors such that the product equals the constant term (45) and the middle term (-34) is split appropriately.
[tex]\[ 5u^2 - 34u + 45 = 0 \][/tex]
Following the factorization process, we find:
[tex]\[ 5u^2 - 34u + 45 = (u - 5)(5u - 9) \][/tex]
Since we substituted [tex]\( u = x^2 \)[/tex], substituting back gives:
[tex]\[ 5x^4 - 34x^2 + 45 = (x^2 - 5)(5x^2 - 9) \][/tex]
### Step 2: Solving for Zeros
The zeros of the polynomial function are the values of [tex]\( x \)[/tex] for which [tex]\( f(x) = 0 \)[/tex]. Therefore, we set each factor equal to zero and solve for [tex]\( x \)[/tex].
1. [tex]\( x^2 - 5 = 0 \)[/tex]
[tex]\[
x^2 = 5 \quad \implies \quad x = \pm\sqrt{5}
\][/tex]
2. [tex]\( 5x^2 - 9 = 0 \)[/tex]
[tex]\[
5x^2 = 9 \quad \implies \quad x^2 = \frac{9}{5} \quad \implies \quad x = \pm\sqrt{\frac{9}{5}} \quad \implies \quad x = \pm\frac{3}{\sqrt{5}} = \pm\frac{3\sqrt{5}}{5}
\][/tex]
### Step 3: Listing All Zeros
Combining the solutions from the steps above, the zeros of the polynomial [tex]\( f(x) = 5x^4 - 34x^2 + 45 \)[/tex] are:
[tex]\[
x = -\sqrt{5}, \quad x = \sqrt{5}, \quad x = -\frac{3\sqrt{5}}{5}, \quad x = \frac{3\sqrt{5}}{5}
\][/tex]
### Summary
The factorized form of the polynomial [tex]\( f(x) = 5x^4 - 34x^2 + 45 \)[/tex] is:
[tex]\[ f(x) = (x^2 - 5)(5x^2 - 9) \][/tex]
The zeros of the polynomial are:
[tex]\[ x = -\sqrt{5}, \quad x = \sqrt{5}, \quad x = -\frac{3\sqrt{5}}{5}, \quad x = \frac{3\sqrt{5}}{5} \][/tex]
