Fourth Degree Polynomials

Step 1 — Find the zeros. Solving \( x^4 = 0 \) gives \( x = 0 \), and the factor \( x^4 \) appears with multiplicity 4.

Step 2 — Multiplicity decides touch/cross. An even multiplicity means the graph touches the x-axis but does not cross it (the sign of the function is the same on both sides of the zero). An odd multiplicity means the graph crosses.

Step 3 — Why so flat? Near \( x = 0 \), the function behaves like \( x^4 \) itself — a fourth-order contact with the x-axis. The curve is nearly horizontal in a small neighborhood of the origin, which is why the graph looks distinctly "flat" there.

Conclusion: the single zero at \( x = 0 \) has even multiplicity 4, so the graph touches without crossing.

Step 1 — Use the given root. Since \( x = 1 \) is an x-intercept, \( (x - 1) \) is a factor:

\[ y = x^4 - x^3 + 0.5x - 0.5 = (x - 1)Q(x) \]

Step 2 — Factor by grouping. Rearrange in pairs:

\[ x^4 - x^3 + 0.5x - 0.5 = x^3(x - 1) + 0.5(x - 1) = (x - 1)(x^3 + 0.5) \]

Step 3 — Solve \( Q(x) = 0 \).

\[ x^3 + 0.5 = 0 \;\Rightarrow\; x^3 = -0.5 \;\Rightarrow\; x = -\sqrt[3]{0.5} \approx -0.7937 \]

Conclusion: the other x-intercept is approximately \( (-0.794,\; 0) \).

Step 1 — Use \( P(2) = 0 \) to find \( a \).

\[ 2^4 - 2(2)^3 - 5(2)^2 + a(2) - 4 = 0 \] \[ 16 - 16 - 20 + 2a - 4 = 0 \;\Rightarrow\; 2a = 24 \;\Rightarrow\; a = 12 \]

The polynomial is now fully determined:

\[ y = x^4 - 2x^3 - 5x^2 + 12x - 4 \]

Step 2 — Determine the multiplicity of \( x = 2 \). Since the graph touches the x-axis at \( x = 2 \), the multiplicity is even. It cannot be 4 (which would account for all four roots and leave no other intercepts — but the graph shows others), so it must be exactly 2.

Thus \( (x - 2)^2 = x^2 - 4x + 4 \) is a factor.

Step 3 — Divide to find \( Q(x) \). Polynomial division gives:

\[ \frac{x^4 - 2x^3 - 5x^2 + 12x - 4}{x^2 - 4x + 4} = x^2 + 2x - 1 \]

Verify by multiplication: \( (x^2 - 4x + 4)(x^2 + 2x - 1) = x^4 - 2x^3 - 5x^2 + 12x - 4 \) ✓

Step 4 — Solve \( x^2 + 2x - 1 = 0 \).

\[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \]

Conclusion: \( a = 12 \), and the other two x-intercepts are exactly \( \bigl(-1 - \sqrt{2},\; 0\bigr) \) and \( \bigl(-1 + \sqrt{2},\; 0\bigr) \), approximately \( (-2.41,\; 0) \) and \( (0.41,\; 0) \).

Step 1 — Factor using the known factor. Dividing \( x^4 + x^3 + 2x^2 + x + 1 \) by \( x^2 + 1 \):

\[ y = (x^2 + 1)(x^2 + x + 1) \]

Verify: \( (x^2 + 1)(x^2 + x + 1) = x^4 + x^3 + x^2 + x^2 + x + 1 = x^4 + x^3 + 2x^2 + x + 1 \) ✓

Step 2 — Check for real zeros.

From \( x^2 + 1 = 0 \): \( x^2 = -1 \) → no real solutions.

From \( x^2 + x + 1 = 0 \): the discriminant is

\[ \Delta = 1^2 - 4(1)(1) = -3 < 0 \]

so this factor also has no real zeros.

Conclusion: the polynomial has no real zeros — its graph never intersects the x-axis. All four roots are complex (in fact, \( \pm i \) and \( \tfrac{-1 \pm i\sqrt{3}}{2} \)).