The task in this image is to find the real roots of the following polynomial equations and then write each polynomial in factored form. Let's go through the equations step by step and solve them.

1. $x^3 - 10x^2 + 32x - 32 = 0$

We can attempt factoring this by synthetic division or trial and error to find the possible roots.

2. $x^3 - 6x^2 + 11x - 6 = 0$

We'll find the roots by factoring this cubic polynomial, possibly using the Rational Root Theorem or synthetic division.

3. $x^3 - 2x^2 + 4x - 8 = 0$

Similar steps will be followed for this cubic polynomial to factor it and find the real roots.

4. $3x^3 - 19x^2 + 33x - 9 = 0$

This cubic polynomial looks more complicated, and we will apply either synthetic division or another method to solve it.

5. $x^4 - 5x^2 + 4 = 0$

This is a quartic equation, but it is a disguised quadratic equation in $x^2$, so we can factor it more easily.

I'll start solving these equations now and provide the roots and factored forms.Here are the real roots and factored forms of the given polynomial equations:

1. $x^3 - 10x^2 + 32x - 32 = 0$

• Real roots: $2, 4$
• Factored form: $(x - 2)^2(x - 4) = 0$

2. $x^3 - 6x^2 + 11x - 6 = 0$

• Real roots: $1, 2, 3$
• Factored form: $(x - 1)(x - 2)(x - 3) = 0$

3. $x^3 - 2x^2 + 4x - 8 = 0$

• Real roots: $2$ (with complex roots)
• Factored form (real part): $(x - 2)(x^2 + 4) = 0$

4. $3x^3 - 19x^2 + 33x - 9 = 0$

• Real roots: $\frac{1}{3}, 3$
• Factored form: $(x - \frac{1}{3})(x - 3)^2 = 0$

5. $x^4 - 5x^2 + 4 = 0$

• Real roots: $-2, -1, 1, 2$
• Factored form: $(x - 2)(x + 2)(x - 1)(x + 1) = 0$

Would you like more details on any of the solutions or methods used?

Related questions:

1. How can synthetic division be used to factor cubic equations?
2. Why do some equations have complex roots?
3. What is the Rational Root Theorem and how is it applied?
4. How does solving quartic equations differ from cubic ones?
5. Can all polynomial equations be factored into real roots?

Tip:

For polynomials with higher degrees, checking for simple integer roots (like ±1, ±2) first can simplify the factorization process.