We're given that $g(x)=x^3-6x^2+11x-6$ has zeros at $x=1,2,3$, and we need to write it in factored form. If a polynomial has zeros at these values, then $(x-1)$, $(x-2)$, and $(x-3)$ are factors. Therefore, $g(x)=(x-1)(x-2)(x-3)$. We can verify by expanding: $(x-1)(x-2)=x^2-3x+2$, then $(x^2-3x+2)(x-3)=x^3-3x^2-3x^2+9x+2x-6=x^3-6x^2+11x-6$. A common sign error is writing $(x+1)$ instead of $(x-1)$ for a zero at $x=1$. Remember: if $a$ is a zero, then $(x-a)$ is the corresponding factor.

The question asks for the degree of p(x) = (x² - 4)(3x³ + 2x - 7) without expanding and its relation to the maximum possible number of real zeros. To determine the degree, note that the first factor has degree 2 and the second has degree 3, so their product has degree 2 + 3 = 5. This degree means the polynomial can have at most 5 real zeros according to the fundamental theorem of algebra, linking the highest power to the potential number of roots. The product's algebraic form allows quick degree identification without full expansion. A key error might be confusing the degree with the product of degrees instead of the sum. A test-taking strategy is to recall that the degree of a product of polynomials is the sum of their degrees, directly informing the maximum zeros.

To find the degree of $p(x)=(x^2-9)(x^3+2x)$, we need the highest power of $x$ when expanded. The highest power comes from multiplying the highest powers in each factor: $x^2 cdot x^3 = x^5$. Therefore, the degree is 5. A common mistake is adding the degrees within each factor (like thinking $x^3+2x$ has degree 4) instead of identifying the highest power term. Remember: degree is determined by the single highest power, not the sum of powers.

The question asks for the expanded standard form of the product (2x - 3)(x² + 4x - 5) after multiplying and combining like terms. To solve, distribute each term in the first binomial to the trinomial: multiply 2x by x² to get 2x³, by 4x to get 8x², and by -5 to get -10x; then multiply -3 by x² to get -3x², by 4x to get -12x, and by -5 to get 15. Combine like terms: 2x³ + (8x² - 3x²) + (-10x - 12x) + 15 simplifies to 2x³ + 5x² - 22x + 15. A common error is mishandling the signs during distribution, such as forgetting the negative from -3, which could lead to incorrect coefficients like in choices B or C. Another mistake might be subtracting instead of adding coefficients, as seen in choice D's negative x² term. When expanding polynomials, always double-check the distribution of negative signs to ensure accurate combination of like terms.

The question asks for a possible factored form of a polynomial that rises to the left, falls to the right, and crosses the x-axis at x=-1, 2, 4. This end behavior indicates an odd degree with negative leading coefficient. With three distinct crossings, the least-degree form is degree 3 with multiplicity 1 at each root: -(x+1)(x-2)(x-4), yielding a -x³ leading term matching the behavior. The factored form connects the roots and leading sign to the graph's intercepts and ends. Common errors include choosing a positive leading coefficient or incorrect roots, like x=1 in choice C. A test-taking strategy is to determine the leading coefficient's sign from end behavior and ensure factors align with given roots.

This question asks for all real zeros of $f(x)=(x+2)(x-5)(x^2+1)$. A polynomial equals zero when any of its factors equals zero. Setting each factor to zero: $x+2=0$ gives $x=-2$, $x-5=0$ gives $x=5$, and $x^2+1=0$ gives $x^2=-1$, which has no real solutions (only complex solutions $x=±i$). Therefore, the real zeros are only $x=-2$ and $x=5$. The key error to avoid is including complex roots like $i$ and $-i$ when asked specifically for real zeros. Remember that $x^2+1$ has no real zeros because squares of real numbers are never negative.

To find the zeros of $f(x)=x^3-6x^2+11x-6$, we need to solve $f(x)=0$. Testing small integer values, we find $f(1)=1-6+11-6=0$, so $(x-1)$ is a factor. Using polynomial division or synthetic division, we get $f(x)=(x-1)(x^2-5x+6)$. Factoring the quadratic: $(x-1)(x-2)(x-3)$, giving zeros at $x=1, 2, 3$. Students often confuse the factors $(x-a)$ with the roots $x=a$, or miss testing $x=1$ as a potential zero. Always verify your zeros by substituting back into the original polynomial.

The question seeks the complete set of zeros for f(x) = x(x - 4)(x + 1)², including multiplicities, as repeated factors indicate multiple instances of the same zero. The zeros are found from each factor: x = 0 from the first (multiplicity 1), x = 4 from the second (multiplicity 1), and x = -1 from the third (multiplicity 2, since squared). Thus, the full list is {-1, -1, 0, 4}, reflecting how multiplicity affects the polynomial's behavior, like touching the x-axis at repeated zeros. Errors often occur by omitting multiplicities, such as listing only distinct zeros like in choice A, or miscounting the repetition as in choices C or D. Another mistake is inverting signs, leading to wrong zeros like -4 or 1. In multiple-choice, verify by plugging zeros back into the factored form to confirm they satisfy f(x) = 0 with the given multiplicities.

We need to factor the expression $(x-4)(x+3)+2(x-4)$. Notice that $(x-4)$ appears in both terms, making it a common factor. Factoring out $(x-4)$ gives us $(x-4)[(x+3)+2] = (x-4)(x+5)$. A common error is to distribute first instead of recognizing the common factor, which makes the problem unnecessarily complex. When you see repeated factors, always look to factor first before expanding.

To find the product $(2x-5)(x^2+3x-4)$, we distribute each term in the first polynomial to every term in the second. This gives us $2x(x^2+3x-4) - 5(x^2+3x-4) = 2x^3+6x^2-8x-5x^2-15x+20$. Combining like terms: $2x^3+(6x^2-5x^2)+(-8x-15x)+20 = 2x^3+x^2-23x+20$. The most common error is sign mistakes when distributing the $-5$, forgetting that $-5 imes -4 = +20$. Always double-check signs when distributing negative terms.