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.

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.

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 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.

To factor $x^2-7x-18$, we need two numbers that multiply to $-18$ and add to $-7$. Testing factor pairs of $-18$: $(-9)(2) = -18$ and $-9 + 2 = -7$, which works! So $x^2-7x-18 = (x-9)(x+2)$. A common error is finding factors that multiply correctly but forgetting to check if they add to the middle coefficient. When the constant term is negative, one factor must be positive and one negative.

The polynomial $k(x)=(x-1)^2(x+4)$ has zeros where each factor equals zero. Setting $(x-1)^2=0$ gives $x=1$ (with multiplicity 2), and $(x+4)=0$ gives $x=-4$. Though $x=1$ appears with multiplicity 2, it's still just one distinct zero. Therefore, $k(x)$ has exactly 2 distinct real zeros: $x=1$ and $x=-4$. Don't confuse multiplicity with the number of distinct zeros—a repeated root counts as one distinct zero.

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.

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 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.

The question seeks the factored form of h(t) = -t² + 6t + 7 that makes the zeros easiest to identify. Factor out the negative: h(t) = -(t² - 6t - 7). The quadratic t² - 6t - 7 factors to (t-7)(t+1), as the factors of -7 summing to -6 are 1 and -7. Thus, h(t) = -(t-7)(t+1), revealing zeros at t=7 and t=-1 directly from the factors. This factored form emphasizes how roots correspond to when each linear factor is zero. A key error is selecting incorrect factor pairs or mishandling the negative sign. A test-taking strategy is to expand the choices back to standard form to verify the match.