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Solved Examples on the Evaluation of Polynomials

Calculators: Calculators for Polynomials
Prerequisites:
(1.) Exponents
(2.) Expressions and Equations
(3.) Relations and Functions

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.

For SAT Students
Any question labeled SAT-C is a question that allows a calculator.
Any question labeled SAT-NC is a question that does not allow a calculator.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.
Unless specified otherwise, any question labeled JAMB is a question from JAMB Physics

For WASSCE Students: Unless specified otherwise:
Any question labeled WASCCE is a question from WASCCE Physics
Any question labeled WASSCE-FM is a question from the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.

(1.) $f(x) = -3x + 7$
Calculate $f(3),\:\:\: f(-5)$

$f(3)$
This means that we have to substitute $3$ for $x$

$ f(x) = -3x + 7 \\[3ex] f(3) = -3(3) + 7 \\[3ex] f(3) = -9 + 7 \\[3ex] f(3) = -2 \\[3ex] $ $f(-5)$
This means that we have to substitute $-5$ for $x$

$ f(x) = -3x + 7 \\[3ex] f(-5) = -3(-5) + 7 \\[3ex] f(-5) = 15 + 7 \\[3ex] f(-5) = 22 $

(2.) $p(x) = -x^2 - 5x + 8$
Calculate $p(0),\:\:\: p(7),\:\:\: p(-7),\:\:\: p(g)$

$p(0)$
This means that we have to substitute $0$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(0) = -1 * 0^2 - 5(0) + 8 \\[3ex] p(0) = -1 * 0 - 0 + 8 \\[3ex] p(0) = 0 - 0 + 8 \\[3ex] p(0) = 8 \\[3ex] $ $p(7)$
This means that we have to substitute $7$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(7) = -1 * 7^2 - 5(7) + 8 \\[3ex] p(7) = -1 * 49 - 35 + 8 \\[3ex] p(7) = -49 - 35 + 8 \\[3ex] p(7) = -76 \\[3ex] $ $p(-7)$
This means that we have to substitute $-7$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(x) = -1 * x^2 - 5x + 8 \\[3ex] p(-7) = -1 * (-7)^2 - 5(-7) + 8 \\[3ex] p(-7) = -1 * 49 + 35 + 8 \\[3ex] p(-7) = -49 + 35 + 8 \\[3ex] p(-7) = -6 \\[3ex] $ $p(g)$
This means that we have to substitute $g$ for $x$

$ p(x) = -x^2 - 5x + 8 \\[3ex] p(g) = -g^2 - 5g + 8 $

(3.) $g(x) = 3x^2 - 3x + 1$
Calculate $g(0),\:\:\: g(-2),\:\:\: g(3),\:\:\: g(-x),\:\:\: g(1 - k)$

$g(0)$
This means that we have to substitute $0$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(0) = 3 * 0^2 - 3 * 0 + 1 \\[3ex] = 3(0) - 0 + 1 \\[3ex] = 0 - 0 + 1 \\[3ex] = 1 \\[3ex] $ $g(-2)$
This means that we have to substitute $-2$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(-2) = 3 * (-2)^2 - 3 * (-2) + 1 \\[3ex] = 3(4) + 6 + 1 \\[3ex] = 12 + 6 + 1 \\[3ex] = 19 \\[3ex] $ $g(3)$
This means that we have to substitute $3$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(3) = 3 * (3)^2 - 3 * (3) + 1 \\[3ex] = 3(9) - 9 + 1 \\[3ex] = 27 - 9 + 1 \\[3ex] = 19 \\[3ex] $ $g(-x)$
This means that we have to substitute $-x$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(-x) = 3 * (-x)^2 - 3 * (-x) + 1 \\[3ex] = 3 * x^2 + 3x + 1 \\[3ex] = 3x^2 + 3x + 1 \\[3ex] $ $g(1-k)$
This means that we have to substitute $(1-k)$ for $x$

$ g(x) = 3x^2 - 3x + 1 \\[3ex] g(1-k) = 3 * (1-k)^2 - 3 * (1-k) + 1 \\[3ex] = 3 * (1-k) * (1-k) - 3(1-k) + 1 \\[3ex] = 3[(1-k)(1-k)] - 3 + 3k + 1 \\[3ex] = 3(1 - k - k + k^2) + 3k - 2 \\[3ex] = 3(1 - 2k + k^2) + 3k - 2 \\[3ex] = 3 - 6k + 3k^2 + 3k - 2 \\[3ex] = 3k^2 - 3k + 1 $

(4.) $g(x) = {\dfrac{2}{3}}x - \dfrac{3}{4}$

Calculate $g(3),\:\:\: g(4),\:\:\: g\left(-\dfrac{1}{2}\right)$

$g(3)$
This means that we have to substitute $3$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 3 - \dfrac{3}{4} \\[5ex] = \dfrac{2}{1} - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{8 - 3}{4} \\[5ex] = \dfrac{5}{4} \\[5ex] $ $g(4)$
This means that we have to substitute $4$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(3) = \dfrac{2}{3} * 4 - \dfrac{3}{4} \\[5ex] = \dfrac{8}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{32}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{32 - 9}{12} \\[5ex] = \dfrac{23}{12} \\[5ex] $ $g\left(-\dfrac{1}{2}\right)$
This means that we have to substitute $-\dfrac{1}{2}$ for $x$

$ g(x) = \dfrac{2}{3}x - \dfrac{3}{4} \\[5ex] g(x) = \dfrac{2}{3} * x - \dfrac{3}{4} \\[5ex] g(4) = \dfrac{2}{3} * -\dfrac{1}{2} - \dfrac{3}{4} \\[5ex] = -\dfrac{1}{3} - \dfrac{3}{4} \\[5ex] = -\dfrac{4}{12} - \dfrac{9}{12} \\[5ex] = \dfrac{-4 - 9}{12} \\[5ex] = -\dfrac{13}{12} $

(5.) $p(m) = m^3$

Calculate $p(3),\:\:\: p(-2),\:\:\: p(-m),\:\:\: p(3y),\:\:\: p(2 + h)$

$p(3)$
This means that we have to substitute $3$ for $m$

$ p(m) = m^3 \\[3ex] p(3) = 3^3 \\[3ex] = 27 \\[3ex] $ $p(-2)$
This means that we have to substitute $(-2)$ for $m$

$ p(m) = m^3 \\[3ex] p(-2) = (-2)^3 \\[3ex] = -8 \\[3ex] $ $p(-m)$
This means that we have to substitute $(-m)$ for $m$

$ p(m) = m^3 \\[3ex] p(3) = (-m)^3 \\[3ex] = (-m)(-m)(-m) \\[3ex] = -m^3 \\[3ex] $ $p(3y)$
This means that we have to substitute $(3y)$ for $m$

$ p(m) = m^3 \\[3ex] p(3y) = (3y)^3 \\[3ex] = 27y^3 \\[3ex] $ $p(2 + h)$
This means that we have to substitute $(2 + h)$ for $m$

$ p(m) = m^3 \\[3ex] p(2 + h) = (2 + h)^3 \\[3ex] = (2 + h)(2 + h)(2 + h) \\[3ex] = (4 + 2h + 2h + h^2)(2 + h) \\[3ex] = (h^2 + 4h + 4)(2 + h) \\[3ex] = (2h^2 + h^3 + 8h + 4h^2 + 8 + 4h) \\[3ex] = h^3 + 6h^2 + 12h + 8 $

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