Question

write the polynomial as the product of linear factors.$h(x) = x^{4} + 10x^{3} + 26x^{2} + 10x + 25LXB0f(x) = square$list all the zeros of the function. (enter your answers as a comma-separated list.)$x = square$resourcesebook

Explanation:

For \(h(x) = x^4 + 10x^3 + 26x^2 + 10x + 25\)

Step1: Group terms for factoring

Group as \((x^4 + 10x^3 + 25x^2) + (x^2 + 10x + 25)\)

Step2: Factor each group

\(x^2(x^2 + 10x + 25) + 1(x^2 + 10x + 25) = (x^2 + 1)(x^2 + 10x + 25)\)

Step3: Factor quadratic to linear

\(x^2 + 1 = (x - i)(x + i)\), \(x^2 +10x +25=(x+5)^2=(x+5)(x+5)\)

Step4: Combine linear factors

\(h(x)=(x - i)(x + i)(x + 5)(x + 5)\)

Step5: Find zeros by setting to 0

Solve \(x - i=0\), \(x + i=0\), \(x + 5=0\)

Step1: Substitute \(u=x^2\)

Let \(u=x^2\), so \(f(u)=u^2 + 34u + 225\)

Step2: Factor quadratic in \(u\)

\(u^2 + 34u + 225=(u + 9)(u + 25)\)

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

\((x^2 + 9)(x^2 + 25)\)

Step4: Factor to linear factors

\(x^2+9=(x-3i)(x+3i)\), \(x^2+25=(x-5i)(x+5i)\)

Step5: Combine linear factors

\(f(x)=(x - 3i)(x + 3i)(x - 5i)(x + 5i)\)

Step6: Find zeros by setting to 0

Solve \(x-3i=0\), \(x+3i=0\), \(x-5i=0\), \(x+5i=0\)

Answer:

\(h(x)=(x + 5)^2(x - i)(x + i)\)
\(x = -5, -5, i, -i\)

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For \(f(x) = x^4 + 34x^2 + 225\)