Polynomials: questions with answers

A non-zero constant such as 7 is $7x^0$, so its degree is 0. The zero polynomial has every coefficient equal to 0, so no power of $x$ is present and there is no highest power to name: its degree is not defined. The two cases differ precisely in whether the constant is zero.

Yes. $x^3$ has degree 3, and $x^3 = 0$ only when $x = 0$, so its only zero is 0. The degree says the graph meets the $x$-axis at most three times; it does not say it must meet it three times.

Yes. The test applies to the POWERS of the variable, which here are 2 and 1, both whole numbers. The coefficients $\sqrt{5}$ and $-\frac{2}{3}$ may be any real numbers, so the definition places no restriction on them. Compare $\sqrt{x} + 2$, which the chapter lists as not a polynomial -- there the root falls on the variable itself, not on a coefficient.

Both 3 and 6 are zeroes of $p$, since $p(x) = (x - 3)(x - 6)$. So $p(3) = 0$ and $p(6) = 0$, and their total is 0. Note that this is not the same as $p(3 + 6)$, which is 18.

The value there is $p(3) = 9 - 18 + 2 = -7$, which is negative, so adding 7 raises it to 0. The amount added is $-p(3)$: the size of $p(3)$, with the sign that cancels it.

The value there is $p(3) = 9 - 3 + 7 = 13$, which is positive, so subtracting 13 leaves the value 0 at $x = 3$. The amount is $p(3)$, not 3 itself and not the constant term.

The zeroes of $p(x)$ are the $x$-coordinates of the points where the graph meets the $x$-axis. This curve meets the $x$-axis at 4 points, so $p(x)$ has 4 zeroes. The point where the curve crosses the $y$-axis is not a zero, and the peaks and troughs are not zeroes either.

The curve lies entirely above the $x$-axis and never reaches it, so there is no value of $x$ at which $p(x) = 0$ and $p(x)$ has no zero. Being a parabola does not guarantee zeroes, and the point where the curve crosses the $y$-axis is not a zero.

The curve comes down to the $x$-axis, touches it at a single point and turns back up without passing through. The zeroes are counted as the DISTINCT points where the graph meets the axis, and there is exactly one such point, so $p(x)$ has one zero. The chapter describes this as the case where the two points of the two-distinct-zeroes picture have coincided into one.

Expanding $a(x-\alpha)(x-\beta)$ gives $ax^2-a(\alpha+\beta)x+a\alpha\beta$. Comparing coefficients gives $\alpha+\beta=-b/a$ and $\alpha\beta=c/a$.

For a cubic $ax^3 + bx^2 + cx + d$, the product of the zeroes is $-\frac{d}{a}$. Here that is $-\frac{8}{2} = -4$; the zeroes are 4, $\frac{1}{2}$ and $-2$.

A quadratic whose only zero is $-2$ has the repeated factor $(x+2)$. Thus $(x+2)^2$ is the required choice.

The sum is 0 and the product is $-4/9$, giving $x^2-4/9$. Multiplying by 45 gives $5(9x^2-4)$ with the same zeroes.

Factorise by taking out the common factor $u$: $5u^2 + 10u = 5u(u + 2)$. Setting each factor to zero gives $u = 0$ and $u = -2$, so the polynomial has 2 zeroes. A student who instead divides the equation by $u$ loses the zero $u = 0$ and finds only 1.

A factor $(x + 3)$ vanishes when $x = -3$, so it is the factor that produces the zero $-3$. The factor $(x - 3)$ would vanish at $+3$ instead: the bracket carries the opposite sign to the zero.