We need a polynomial that crosses the $x$-axis at $x=-2$ and $x=3$, and touches (bounces off) at $x=1$. When a polynomial crosses the $x$-axis at a point, the corresponding factor has odd multiplicity (usually 1); when it touches without crossing, the factor has even multiplicity (usually 2). Therefore, we need factors $(x+2)^1$, $(x-3)^1$, and $(x-1)^2$. The polynomial with least degree and leading coefficient 1 is $(x+2)(x-3)(x-1)^2$. This gives the correct crossing/touching behavior at each intercept. Remember: odd multiplicity means crossing, even multiplicity means touching.

We need a polynomial that touches (doesn't cross) at $x=-2$ and crosses at $x=1$. Touching without crossing indicates even multiplicity, while crossing indicates odd multiplicity. At $x=-2$, we need factor $(x+2)^2$ (even multiplicity); at $x=1$, we need factor $(x-1)^1$ (odd multiplicity). The polynomial with least degree and positive leading coefficient is $(x+2)^2(x-1)$. This has degree 3 and expands with positive leading coefficient. The key insight is that the graph's behavior at each intercept determines the required multiplicity of the corresponding factor.

We need to find the zeros of $r(x)=x^2+6x+9$ and determine how the graph meets the x-axis. First, we can factor this quadratic: we need two numbers that multiply to 9 and add to 6, which are both 3. So $r(x) = (x+3)(x+3) = (x+3)^2$. Setting this equal to zero: $(x+3)^2 = 0$ gives $x = -3$ with multiplicity 2. Since the zero has even multiplicity, the graph touches but doesn't cross the x-axis at $x=-3$. This is a perfect square trinomial, which always results in a repeated root where the parabola just touches the x-axis.

We need to expand $q(x)=(x-7)(x^2+7x+49)$, which follows the difference of cubes pattern. Distributing: $x(x^2+7x+49) - 7(x^2+7x+49) = x^3 + 7x^2 + 49x - 7x^2 - 49x - 343$. Combining like terms: $x^3 + (7-7)x^2 + (49-49)x - 343 = x^3 - 343$. This is indeed $x^3 - 7^3$, confirming the difference of cubes pattern: $(a-b)(a^2+ab+b^2) = a^3-b^3$. The middle terms cancel out, which is characteristic of this special factorization. Recognizing special patterns like difference of cubes can save time and reduce errors.

The polynomial $p(x)=x^6-3x^2+10$ is already in standard form, and we need to identify its degree. The degree is the highest power of x with a non-zero coefficient, which is 6 (from the $x^6$ term). A polynomial of degree 6 can have at most 6 real zeros, which means its graph can have at most 6 x-intercepts. Note that this polynomial has no odd-degree terms ($x^5$, $x^3$, $x$), making it an even function. Don't be misled by the missing terms; the degree is still determined by the highest power present.

To factor $m(x)=2x^4-8x^3+6x^2$, we first identify the greatest common factor (GCF) of all terms. Each term has a factor of $x^2$: $2x^4 = 2x^2 \cdot x^2$, $-8x^3 = -8x^2 \cdot x$, $6x^2 = 6x^2 \cdot 1$. The coefficients 2, -8, and 6 have GCF of 2. Therefore, the overall GCF is $2x^2$. Factoring this out: $m(x) = 2x^2(x^2 - 4x + 3)$. We can verify by expanding: $2x^2(x^2 - 4x + 3) = 2x^4 - 8x^3 + 6x^2$ ✓. Always factor out the GCF first when working with polynomials, as it simplifies further factoring steps.

We need to expand $ (x^2+2x-3)(x-4) $ by distributing each term in the first polynomial to $ (x-4) $. First, distribute $x^2$: $x^2(x) - x^2(4) = x^3 - 4x^2$. Next, distribute $2x$: $2x(x) - 2x(4) = 2x^2 - 8x$. Finally, distribute $-3$: $-3(x) - (-3)(4) = -3x + 12$. Combining all terms: $x^3 - 4x^2 + 2x^2 - 8x - 3x + 12 = x^3 - 2x^2 - 11x + 12$. The key is to carefully track signs, especially when distributing negative terms. When multiplying polynomials, organize your work to avoid missing terms or making sign errors.

The polynomial has degree 4 with zeros at $x=-1$ (multiplicity 1) and $x=2$ (multiplicity 3), and leading coefficient 5. Since the multiplicities must sum to the degree: 1 + 3 = 4 ✓. The factored form is $k(x) = 5(x-(-1))^1(x-2)^3 = 5(x+1)(x-2)^3$. The factor $(x+1)$ corresponds to the zero at $x=-1$, and $(x-2)^3$ corresponds to the zero at $x=2$ with multiplicity 3. The graph will cross at $x=-1$ (odd multiplicity) and have a flattened crossing at $x=2$ (odd multiplicity ≥ 3). When constructing polynomials from zeros and multiplicities, ensure the sum of multiplicities equals the degree.

The graph crosses the x-axis at $x=-3$ (indicating odd multiplicity) and touches/bounces at $x=1$ (indicating even multiplicity). With a positive leading coefficient, the simplest polynomial matching this behavior is $(x+3)(x-1)^2$. The factor $(x+3)$ has multiplicity 1 (odd), so the graph crosses at $x=-3$. The factor $(x-1)^2$ has multiplicity 2 (even), so the graph touches but doesn't cross at $x=1$. The positive leading coefficient (which is 1 when expanded) ensures the polynomial goes to $+infty$ as $x o infty$. Remember that crossing behavior at x-intercepts is determined by the multiplicity: odd multiplicities cross, even multiplicities touch.

To find which value is a zero of $f(x)=x^3-6x^2+11x-6$, we need to test each option by substitution. Testing $x=1$: $f(1) = 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0$. Since $f(1) = 0$, we know $x=1$ is a zero. We can verify by factoring: since $(x-1)$ is a factor, we can use polynomial division to find $f(x) = (x-1)(x^2-5x+6) = (x-1)(x-2)(x-3)$. The zeros are $1$, $2$, and $3$. For polynomial zero problems, the rational root theorem can help narrow down candidates to test.