Polynomials and Their Graphs: Practice

The leading term of a product is the product of the leading terms, since only that pairing reaches the top degree.

$(3x^2)(x^3)(x) = 3x^6$

Degrees add to $2 + 3 + 1 = 6$ and the coefficients multiply to $3 \cdot 1 \cdot 1 = 3$. The $-5$ never touches the leading term.

The sum of a polynomial's coefficients is its value at $x = 1$, so there is no need to expand.

$(2 \cdot 1 - 1)^4 = 1^4 = 1$

Every coefficient of the expansion, positive and negative, nets out to $1$.

Five real zeros force the degree to be at least $5$. But both ends rising means the far ends agree, which happens only for even degree.

$n \ge 5 \text{ and } n \text{ even} \;\Rightarrow\; n \ge 6$

Those ceilings only rule degrees out, so check that $6$ is reached: $x^2(x - 1)(x - 2)(x - 3)(x - 4)$ has degree $6$, rises at both ends, and has exactly the five zeros $0$, $1$, $2$, $3$, $4$. So the least possible degree is $6$.

The coefficient sum is the value at $1$, so set $p(1) = 0$.

$p(1) = 2 - 7 + k - 5 = k - 10 = 0$

Solving gives $k = 10$.

The graph of $y = x^3$ falls on the left and rises on the right: odd degree, positive leading coefficient. Match those two features.

$5x^7 - x^2: \quad n = 7 \text{ (odd)}, \qquad a_n = 5 > 0$

The first and last options have even degree, and $-4x^5 + 1$ has odd degree but a negative leading coefficient, which reverses both ends.

Between two neighboring $x$-intercepts the unbroken graph leaves the axis and must come back, so it turns around at least once in between.

$6 \text{ zeros} \;\Rightarrow\; \text{at least } 6 - 1 = 5 \text{ turning points}$

Five gaps between six zeros force at least five turning points.

Squaring multiplies $f$ by itself, and degrees add in a product.

$\deg\left[f(x)\right]^2 = 3 + 3 = 6$

Adding $f$, a degree-$3$ polynomial, cannot touch the degree-$6$ leading term, so the sum has degree $6$.

Collect only the pairings whose exponents add to $3$: $x^2$ times $-3x$, and $2x$ times $x^2$.

$(x^2)(-3x) + (2x)(x^2) = -3x^3 + 2x^3 = -x^3$

So the coefficient of $x^3$ is $-1$; no other pairing produces a cube.

A higher power eventually outgrows any constant multiple of a lower power. Compare the top two directly:

$0.01x^5 > 1000x^4 \iff x > 100{,}000$

Beyond that threshold the fifth power leads, and it dwarfs the cubic and the constant even sooner. Coefficients set the threshold; the exponent decides the winner.

Falling left and rising right needs an odd degree with a positive leading coefficient, and the constant term must be $-6$.

$x^3 + 2x - 6: \quad n = 3 \text{ (odd)}, \quad a_n = 1 > 0, \quad f(0) = -6$

The first option has a negative lead (ends reversed), the second has even degree (ends agree), and the last has $y$-intercept $0$.

Substituting $-x$ flips the sign of every odd-power term and fixes every even-power term, so $p(x) - p(-x)$ is twice the odd-power part of $p$. The identity forces that part to vanish:

$p(x) - p(-x) = 2\left(a_1 x + a_3 x^3 + a_5 x^5 + \cdots\right) = 0$

So every odd coefficient is $0$, including the coefficient of $x^3$. The even coefficients are unrestricted, which kills the other options: $p(x) = x^2$ has a nonzero $x^2$ coefficient and a real zero, and $p(x) = x^2 + 1$ has a nonzero constant term.

Substitute $x = 2$ and solve for $a$.

$f(2) = 8a + 2(4) - 5 = 8a + 3 = 19$

So $8a = 16$ and $a = 2$.