Polynomials: Competency Practice

Parabolic trajectories, apex coordinates, and analytical zero-coefficient relations.

6 problems·20–25 min·★★★★☆

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Problem 1
polynomials·★★★★★
The vertical height $h$ (in metres) of a decorative water fountain stream at horizontal distance $x$ (in metres) from the nozzle is modeled by the quadratic polynomial $h(x) = 4x - \frac{1}{2}x^2$. (i) Find the zeroes of $h(x)$ and state their physical significance. (ii) Find the maximum height reached by the water stream above ground. (iii) Calculate the height of the stream at horizontal distance $2\text{ m}$ from the nozzle.
▶Answer
(i) $x = 0$ and $x = 8$ (nozzle point and landing point at ground level), (ii) $8\text{ m}$, (iii) $6\text{ m}$
▶Step-by-step solution
(i) Setting $h(x) = 0$ gives $x = 0$ (nozzle launch origin) and $x = 8\text{ m}$ (landing point where water hits ground level).
(ii) By parabolic symmetry, maximum height occurs at the midpoint $x = 4\text{ m}$, giving $h(4) = 8\text{ m}$.
(iii) At $x = 2\text{ m}$, the stream height is $h(2) = 6\text{ m}$.
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Problem 2
Factorisation·★★★★☆
Factorise $3x^2 + 18x + 15$, find its zeroes, and verify the coefficient relations.
▶Answer
The zeroes are $-1$ and $-5$.
▶Step-by-step solution
Factorise the polynomial: $3x^2 + 18x + 15$ = $3(x + 1)(x + 5)$.
Set each factor equal to $0$; this gives $x = -1$ and $x = -5$.
Sum of zeroes: $-6$; from coefficients, $-b/a = -6$.
Product of zeroes: $5$; from coefficients, $c/a = 5$.
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Problem 3
Quadratic Polynomial Construction·★★★☆☆
Find a quadratic polynomial with zeroes adding to $8$ and multiplying to $12$; also find the zeroes.
▶Answer
One suitable polynomial is $t^2 - 8t + 12$; its zeroes are $2$ and $6$.
▶Step-by-step solution
For zeroes with sum $8$ and product $12$, use $t^2 - (sum)t + product$.
So the polynomial is $t^2 - 8t + 12$.
Factorising gives $t^2 - 8t + 12$ = $(t - 2)(t - 6)$.
Hence the zeroes are $2$ and $6$.
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Problem 4
Graphs of Polynomials·★★★☆☆
Write true or false, with a reason: If the graph of a quadratic polynomial intersects the x-axis at only one point, its two zeroes cannot be equal.
upward-opening parabola touching the x-axis once
▶Answer
False. A quadratic graph that touches the x-axis has two equal zeroes.
▶Step-by-step solution
The zeroes of a polynomial are the x-coordinates where its graph meets the x-axis.
A quadratic with equal zeroes has a parabola tangent to the x-axis.
That graph has one x-intercept representing the repeated zero.
Therefore one x-axis intersection is consistent with two equal zeroes.
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Problem 5
Factorisation·★★★★★
Factorise $10y^2 + 12\sqrt{5}y + 16$, find its zeroes, and verify the coefficient relations.
▶Answer
The zeroes are $-\frac{4\sqrt{5}}{5}$ and $-\frac{2\sqrt{5}}{5}$.
▶Step-by-step solution
Factorise the polynomial: $10y^2 + 12\sqrt{5}y + 16$ = $2(\sqrt{5}y + 4)(\sqrt{5}y + 2)$.
Set each factor equal to $0$; this gives $y = -\frac{4\sqrt{5}}{5}$ and $y = -\frac{2\sqrt{5}}{5}$.
Sum of zeroes: $-\frac{6\sqrt{5}}{5}$; from coefficients, $-b/a = -\frac{6\sqrt{5}}{5}$.
Product of zeroes: $\frac{8}{5}$; from coefficients, $c/a = \frac{8}{5}$.
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Problem 6
Quadratic Polynomial Construction·★★★★☆
Find a quadratic polynomial with zeroes adding to $-\frac{7\sqrt{5}}{5}$ and multiplying to $\frac{12}{5}$; also find the zeroes.
▶Answer
One suitable polynomial is $x^2 + \frac{7\sqrt{5}}{5}x + \frac{12}{5}$; its zeroes are $-\frac{4\sqrt{5}}{5}$ and $-\frac{3\sqrt{5}}{5}$.
▶Step-by-step solution
For zeroes with sum $-\frac{7\sqrt{5}}{5}$ and product $\frac{12}{5}$, use $x^2 - (sum)x + product$.
So the polynomial is $x^2 + \frac{7\sqrt{5}}{5}x + \frac{12}{5}$.
Factorising gives $x^2 + \frac{7\sqrt{5}}{5}x + \frac{12}{5}$ = $(x + \frac{4\sqrt{5}}{5})(x + \frac{3\sqrt{5}}{5})$.
Hence the zeroes are $-\frac{4\sqrt{5}}{5}$ and $-\frac{3\sqrt{5}}{5}$.
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