Polynomials and Their Graphs: Additional Practice (Set 1)

Take the leading piece of each factor: $2x^3$ from the first, and $(-x)^2 = x^2$ from $(4 - x)^2$, then apply the outer minus sign.

$-(2x^3)(x^2) = -2x^5$

Degrees add to $3 + 2 = 5$; the squared factor contributes a positive $x^2$ because squaring erases the sign of $-x$.

Multiply the leading pieces of the factors: $(-x)(x)(x^2)$.

$(-x)(x)(x^2) = -x^4$

The degree $4$ is even, so the two ends agree, and the leading coefficient $-1$ points them down: both ends fall.

The coefficient sum is the value at $x = 1$, so evaluate the compact form instead of expanding.

$(1 - 3 + 1)^3 = (-1)^3 = -1$

All the expansion's coefficients net out to $-1$.

Take the powers of $\tfrac{1}{2}$ first: $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ and $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

$f\left(\tfrac{1}{2}\right) = 2 \cdot \tfrac{1}{8} - 3 \cdot \tfrac{1}{4} + 4 \cdot \tfrac{1}{2} - 5 = \tfrac{1}{4} - \tfrac{3}{4} + 2 - 5$

The fractions combine to $-\tfrac{1}{2}$, and $-\tfrac{1}{2} + 2 - 5 = -\tfrac{7}{2}$.

Expand the square first: $(x + 2)^2 = x^2 + 4x + 4$. Then collect only the pairings with $(x - 1)$ whose exponents add to $2$:

$(x^2)(-1) + (4x)(x) = -x^2 + 4x^2 = 3x^2$

The full product is $x^3 + 3x^2 - 4$, confirming the coefficient $3$.

Far from the origin the leading term $4x^3$ dictates the sign, and $-1000$ is far out: the threshold from the leading-term argument is $|x| \ge \max\left(1, \tfrac{2(7 + 3)}{4}\right) = 5$.

$x = -1000: \quad 4x^3 = 4(-1000)^3 < 0$

An odd power keeps the sign of its base, so the leading term, and with it $f(-1000)$, is negative. This is exactly why an odd positive-lead graph falls on the far left.

Test by evaluation, looking for an input whose sixth power is $64$.

$p(2) = 2^6 - 64 = 64 - 64 = 0$

So $2$ is a zero. The others overshoot wildly: $4^6 = 4096$, and $8^6$ and $64^6$ are larger still.

Feed in any $x > 0$: every term $c\,x^k$ has a positive coefficient and a positive power, so every term is positive.

$x > 0 \;\Rightarrow\; f(x) > 0$

A sum of positive numbers is never $0$, so no positive input can be a zero: the count is exactly $0$. Any real zeros such a polynomial has must be negative, since $x = 0$ fails too, $f(0)$ being the positive constant term.

Expand and watch the top terms cancel.

$(x^2 + 1)(x^2 + 2) - x^4 = (x^4 + 3x^2 + 2) - x^4 = 3x^2 + 2$

Subtraction can destroy leading terms, so the degree must be read after simplifying: it is $2$, not $4$.

Only the leading term $-6x^5$ matters far out. For a large negative $x$, the odd power $x^5$ is negative, and the coefficient $-6$ flips that sign.

$x \text{ large negative}: \quad -6x^5 = (-6)(\text{negative}) > 0$

So the outputs grow past any bound: the graph rises on the far left, as it must for an odd degree with a negative lead.

A zero makes the output vanish, so set $f(-2) = 0$ with $(-2)^3 = -8$.

$f(-2) = -8 - 2c + 10 = 2 - 2c = 0$

Solving gives $c = 1$.

Polynomial graphs are smooth: they bend gradually and never crease.

$\text{sharp corner} \;\Rightarrow\; \text{not a polynomial graph}$

The others all happen: $x(x-1)(x-2)(x-3)(x-4)$ crosses five times, $x^2 + 1$ stays above the axis, and $-x^2$ falls on both ends. A corner like the one on $y = |x|$ is the giveaway.