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:

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

Step1: Recognize reciprocal polynomial

Notice \(h(x)\) is a palindromic polynomial: coefficients read same forwards/backwards. Divide by \(x^2\):
\(x^2 + 10x + 26 + \frac{10}{x} + \frac{25}{x^2} = 0\)

Step2: Substitute \(y = x + \frac{1}{x}\)

Rewrite as \(y^2 + 10y + 24 = 0\), since \(x^2+\frac{1}{x^2}=y^2-2\)

Step3: Solve quadratic for \(y\)

\((y+4)(y+6)=0 \implies y=-4, y=-6\)

Step4: Solve for \(x\) from \(y=x+\frac{1}{x}\)

For \(y=-4\): \(x^2+4x+1=0 \implies x=-2\pm\sqrt{3}\)
For \(y=-6\): \(x^2+6x+1=0 \implies x=-3\pm\sqrt{8}=-3\pm2\sqrt{2}\)

Step5: Write linear factors

Each root \(a\) gives factor \((x-a)\)

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

Rewrite as quadratic: \(y^2+34y+225=0\)

Step2: Factor quadratic for \(y\)

\((y+9)(y+25)=0 \implies y=-9, y=-25\)

Step3: Solve for \(x\) from \(y=x^2\)

\(x^2=-9 \implies x=\pm3i\); \(x^2=-25 \implies x=\pm5i\)

Step4: Write linear factors

Each root \(a\) gives factor \((x-a)\)

Answer:

\(h(x)=(x+2-\sqrt{3})(x+2+\sqrt{3})(x+3-2\sqrt{2})(x+3+2\sqrt{2})\)
Zeros: \(x=-2+\sqrt{3}, -2-\sqrt{3}, -3+2\sqrt{2}, -3-2\sqrt{2}\)

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