[-/2 Points] DETAILS MY NOTES SPRECALC8 3.2.035.MI. Factor the polynomial and use the factored form to find the zeros. (Enter your answers as a comma-separated list. Enter all answers using the appropriate multiplicities.) \[ P=\square \text { ASK YOUR TEACHER } \] PRAC \( x=x^{4}-5 x^{3}+4 x^{2} \) Sketch the graph. \( y \)

To factor the polynomial \( P(x) = x^4 - 5x^3 + 4x^2 \) and find its zeros, we will follow these steps: 1. **Factor out the greatest common factor (GCF)**. 2. **Factor the remaining polynomial**. 3. **Find the zeros from the factored form**. 4. **Sketch the graph based on the zeros and their multiplicities**. ### Step 1: Factor out the GCF The GCF of the terms in \( P(x) \) is \( x^2 \). We can factor this out: \[ P(x) = x^2(x^2 - 5x + 4) \] ### Step 2: Factor the remaining polynomial Next, we need to factor the quadratic \( x^2 - 5x + 4 \). We can look for two numbers that multiply to \( 4 \) (the constant term) and add to \( -5 \) (the coefficient of \( x \)). The numbers \( -4 \) and \( -1 \) satisfy these conditions. Thus, we can factor the quadratic as follows: \[ x^2 - 5x + 4 = (x - 4)(x - 1) \] Putting it all together, we have: \[ P(x) = x^2(x - 4)(x - 1) \] ### Step 3: Find the zeros To find the zeros, we set \( P(x) = 0 \): \[ x^2(x - 4)(x - 1) = 0 \] This gives us the following equations: 1. \( x^2 = 0 \) → \( x = 0 \) (with multiplicity 2) 2. \( x - 4 = 0 \) → \( x = 4 \) (with multiplicity 1) 3. \( x - 1 = 0 \) → \( x = 1 \) (with multiplicity 1) Thus, the zeros of the polynomial are: \[ 0 \text{ (multiplicity 2)}, 1 \text{ (multiplicity 1)}, 4 \text{ (multiplicity 1)} \] ### Final Answer for Zeros The zeros, including their multiplicities, are: \[ 0, 0, 1, 4 \] ### Step 4: Sketch the graph To sketch the graph of \( P(x) \): - The zero at \( x = 0 \) has a multiplicity of 2, which means the graph will touch the x-axis at this point and turn around. - The zero at \( x = 1 \) has a multiplicity of 1, so the graph will cross the x-axis at this point. - The zero at \( x = 4 \) also has a multiplicity of 1, so the graph will cross the x-axis at this point as well. The general shape of the graph will start from positive infinity (as \( x \to -\infty \)), touch the x-axis at \( x = 0 \), cross at \( x = 1 \), and cross again at \( x = 4 \), eventually going to positive infinity (as \( x \to +\infty \)). This completes the solution. If you need a visual representation or further assistance, please let me know!