Q. \(x^4 – 10x^3 + 35x^2 – 50x + 25\) factorization.

Write the polynomial to factor:

\[x^{4} – 10x^{3} + 35x^{2} – 50x + 25\]

Try to see if it is a perfect square of a quadratic. Assume

\[\bigl(x^{2} + p x + q\bigr)^{2} = x^{4} – 10x^{3} + 35x^{2} – 50x + 25.\]

Expand the left-hand side:

\[
\bigl(x^{2} + p x + q\bigr)^{2}
= x^{4} + 2p\,x^{3} + \bigl(p^{2} + 2q\bigr)x^{2} + 2pq\,x + q^{2}.
\]

Equate coefficients of like powers of x between the expanded form and the given polynomial:

• Coefficient of x^{3}: \[2p = -10 \implies p = -5.\]
• Coefficient of x^{2}: \[p^{2} + 2q = 35.\] Substitute p = -5 to get \[25 + 2q = 35 \implies 2q = 10 \implies q = 5.\]
• Verify the remaining coefficients:

  Coefficient of x: \[2pq = 2(-5)(5) = -50\] matches.

  Constant term: \[q^{2} = 5^{2} = 25\] matches.

Therefore the polynomial is the square of the quadratic found:

\[\boxed{\bigl(x^{2} – 5x + 5\bigr)^{2}}\]

If desired, factor further into linear factors over the real numbers by finding the roots of x^{2} – 5x + 5. Compute the discriminant:

\[\Delta = (-5)^{2} – 4\cdot 1\cdot 5 = 25 – 20 = 5.\]

The roots are

\[
x = \frac{5 \pm \sqrt{5}}{2}.
\]

Hence the full factorization into linear real factors is

\[
\boxed{\biggl(x – \frac{5 + \sqrt{5}}{2}\biggr)^{2}\biggl(x – \frac{5 – \sqrt{5}}{2}\biggr)^{2}}.
\]