Add, Subtract, Multiply, and Divide Polynomials

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.
Use at least two approaches as applicable.
Show all work.

For questions on the division of polynomials:
Unless otherwise specified, write your answers using any of the approved formats.
Check your solution using the Division Algorithm

(1.) Multiply $(7 + 5c)(7 - 5c)$

We can multiply the binomials using at least two approaches.
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;FOIL\;\;Method} \\[3ex] (7 + 5c)(7 - 5c) \\[3ex] First:\;\; 7(7) = 49 \\[3ex] Outer:\;\; 7(-5c) = -35c \\[3ex] Inner:\;\; 5c(7) = 35c \\[3ex] Last:\;\; 5c(-5c) = -25c^2 \\[3ex] \implies \\[3ex] = 49 - 35c + 35c - 25c^2 \\[3ex] = 49 - 25c^2 \\[5ex] \underline{Second\;\;Approach:\;\;Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (7 + 5c)(7 - 5c) \\[3ex] = 7^2 - (5c)^2 \\[3ex] = 49 - 5^2c^2 \\[3ex] = 49 - 25c^2 $

(3.) Multiply $12m^4(7m + 9m^2 + 3)$

We can multiply the binomials using at least three approaches.
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;Horizontal\;\;Method} \\[3ex] 12m^4(7m + 9m^2 + 3) \\[3ex] 12m^4(7m) = 84m^5 \\[3ex] 12m^4(9m^2) = 108m^6 \\[3ex] 12m^4(3) = 36m^4 \\[3ex] \implies \\[3ex] = 84m^5 + 108m^6 + 36m^4 \\[3ex] = 108m^6 + 84m^5 + 36m^4 \\[5ex] \underline{Second\;\;Approach:\;\;Box\;\;Method} \\[3ex] Standard\;\;Form\;\;for\;\;2nd\;\;Polynomial = 9m^2 + 7m + 3 \\[3ex] \begin{array} {c|c|} &9m^2 & +7m & +3 \hspace{3em} \\ \hline 12m^4 & 108m^6 & +84m^5 & +36m^4 \\ \hline \end{array} \\[3ex] Product = 108m^6 + 84m^5 + 36m^4 \\[5ex] \underline{Third\;\;Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 9m^2 + 7m + 3 \\ * \hspace{7em} \\ \hspace{4em} 12m^4 \\ \hline 108m^6 + 84m^5 + 36m^4 \\ \hline \end{array} $

(5.) Simplify $(-5p^2 + 2p - 7) + (-3p^2 + 3p)$

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \underline{Horizontal\;\;Method} \\[3ex] (-5p^2 + 2p - 7) + (-3p^2 + 3p) \\[3ex] 1(-5p^2 + 2p - 7) + 1(-3p^2 + 3p) \\[3ex] -5p^2 + 2p - 7 + -3p^2 + 3p \\[3ex] -5p^2 + 2p - 7 - 3p^2 + 3p \\[3ex] -5p^2 - 3p^2 + 2p + 3p - 7 \\[3ex] -8p^2 + 5p - 7 \\[5ex] \underline{Vertical\;\;Method} \\[3ex] \begin{array}{c} -5p^2 + 2p - 7 \\ +\hspace{7em} \\ -3p^2 + 3p + 0 \\ \hline -8p^2 + 5p - 7 \\ \hline \end{array} $

(6.) Simplify $(-5p^2 + 2p - 7) - (-3p^2 + 3p)$

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \underline{Horizontal\;\;Method} \\[3ex] (-5p^2 + 2p - 7) - (-3p^2 + 3p) \\[3ex] 1(-5p^2 + 2p - 7) - 1(-3p^2 + 3p) \\[3ex] -5p^2 + 2p - 7 + 3p^2 - 3p \\[3ex] -5p^2 + 2p - 7 + 3p^2 - 3p \\[3ex] -5p^2 + 3p^2 + 2p - 3p - 7 \\[3ex] -2p^2 - p - 7 \\[5ex] \underline{Vertical\;\;Method} \\[3ex] \begin{array}{c} -5p^2 + 2p - 7 \\ -\hspace{7em} \\ -3p^2 + 3p + 0 \\ \hline -2p^2 - p - 7 \\ \hline \end{array} $

(7.) ACT The lengths of 2 adjacent sides of a rectangle are represented by $x + 2$ feet and $2x + 7$ feet.
In terms of x, what is the area, in square feet, of the rectangle?

$ A.\;\; 6x + 18 \\[3ex] B.\;\; 2x^2 + 14 \\[3ex] C.\;\; 2x^2 + 9x + 14 \\[3ex] D.\;\; 2x^2 + 11x + 9 \\[3ex] E.\;\; 2x^2 + 11x + 14 \\[3ex] $

$ Area\;\;of\;\;a\;\;rectangle = length * width \\[3ex] So,\;\;multiply\;\;the\;\;two\;\;adjacent\;\;sides \\[3ex] In\;\;other\;\;words,\;\;multiply\;\;the\;\;two\;\;binomials \\[3ex] Area \\[3ex] = (x + 2)(2x + 7) \\[5ex] \underline{FOIL\;\;Method} \\[3ex] First:\;\; x(2x) = 2x^2 \\[3ex] Outer:\;\; x(7) = 7x \\[3ex] Inner:\;\; 2(2x) = 4x \\[3ex] Last:\;\; 2(7) = 14 \\[3ex] = 2x^2 + 7x + 4x + 14 \\[3ex] = 2x^2 + 11x + 14\;square\;\;feet $

(8.) JAMB Find the remainder when $3x^3 + 5x^2 - 11x + 4$ is divided by $x + 3$

$ A.\;\; 1 \\[3ex] B.\;\; -4 \\[3ex] C.\;\; 4 \\[3ex] D.\;\; -1 \\[3ex] $

We can solve the question using at least two approaches.
Use any approach you prefer.

$ \underline{First\;\;Approach:\;\;Remainder\;\;Theorem} \\[3ex] Factor:\;\; x + 3 \\[3ex] Zero:\;\; x + 3 = 0 \\[3ex] x = -3 \\[3ex] f(x) = 3x^3 + 5x^2 - 11x + 4 \\[3ex] Remainder = f(-3) \\[3ex] = 3(-3)^3 + 5(-3)^2 - 11(-3) + 4 \\[3ex] = 3(-27) + 5(9) + 33 + 4 \\[3ex] = -81 + 45 + 33 + 4 \\[3ex] = 1 \\[5ex] \underline{Second\;\;Approach:\;\;Long\;\;Division} \\[3ex] \begin{array}{c|c} &3x^2 - 4x + 1 \hspace{3em} \\ \hline x + 3 & 3x^3 + 5x^2 - 11x + 4 \\ &- \hspace{10em} \\ &3x^3 + 9x^2 \hspace{4.5em} \\ \hline &-4x^2 - 11x \\ &- \hspace{8em} \\ &-4x^2 - 12x \\ \hline &\hspace{6em} x + 4 \\ &\hspace{2em} - \\ &\hspace{6em} x + 3 \\ \hline &\hspace{8em} 1 \\ \end{array} \\[5ex] Remainder = 1 $

(9.) Peform the indicated operations.

$ (a.)\;\; (10p^4 + 11) - (11p^4 + 13 + 16p^2) \\[3ex] (b.)\;\; (10x^4 - 16) + (12 - 6x^3 + 11x^4) \\[3ex] (c.)\;\; (20v^2 - 9v^3) - (7v^3 - 10v^4 - 14v^2) \\[3ex] $

We can solve these questions using at least two approaches.
Use any approach you prefer.

$ (a.) \\[3ex] (10p^4 + 11) - (11p^4 + 13 + 16p^2) \\[3ex] \underline{First\:\:Approach:\;\;Horizontal\;\;Method} \\[3ex] (10p^4 + 11) - (11p^4 + 13 + 16p^2) \\[3ex] = 1(10p^4 + 11) - 1(11p^4 + 13 + 16p^2) \\[3ex] Distributive\:\:Property \\[3ex] = 10p^4 + 11 - 11p^4 - 13 - 16p^2 \\[3ex] Collect\:\:like\:\:terms \\[3ex] = 10p^4 - 11p^4 - 16p^2 + 11 - 13 \\[3ex] = -p^4 - 16p^2 - 2 \\[5ex] \underline{Second\:\:Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 10p^4 + 0p^2 + 11 \\ -\hspace{8em} \\ ~11p^4 + 16p^2 + 13 \\ \hline -p^4 - 16p^2 - 2 \\ \hline \\[5ex] \end{array} \\[5ex] (b.) \\[3ex] (10x^4 - 16) + (12 - 6x^3 + 11x^4) \\[3ex] \underline{First\:\:Approach:\;\;Horizontal\;\;Method} \\[3ex] 10x^4 - 16 + 12 - 6x^3 + 11x^4 \\[3ex] 10x^4 + 11x^4 - 6x^3 - 16 + 12 \\[3ex] 21x^4 - 6x^3 - 4 \\[5ex] \underline{Second\:\:Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 10x^4 + 0x^3 - 16 \\ +\hspace{8em} \\ 11x^4 - 6x^3 + 12 \\ \hline 21x^4 - 6x^3 - 4~ \\ \hline \\[5ex] \end{array} \\[5ex] (c.) \\[3ex] (20v^2 - 9v^3) - (7v^3 - 10v^4 - 14v^2) \\[3ex] \underline{First\:\:Approach:\;\;Horizontal\;\;Method} \\[3ex] 20v^2 - 9v^3 - 7v^3 + 10v^4 + 14v^2 \\[3ex] 10v^4 - 9v^3 - 7v^3 + 20v^2 + 14v^2 \\[3ex] 10v^4 - 16v^3 + 34v^2 \\[5ex] \underline{Second\:\:Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} ~~0v^4 - 9v^3 + 20v^2 \\ -\hspace{10em} \\ -10v^4 + 7v^3 - 14v^2 \\ \hline 10v^4 - 16v^3 + 34v^2 \\ \hline \end{array} $

(10.) ACT Given constants c, d, m and n such that $x^2 + mx + c$ has factors of (x + 2) and (x + 4) and $x^2 + nx + d$ has factors of (x + 3) and (x + 7), what is mn?

$ A.\;\; 16 \\[3ex] B.\;\; 18 \\[3ex] C.\;\; 29 \\[3ex] D.\;\; 60 \\[3ex] E.\;\; 168 \\[3ex] $

$ (x + 2)(x + 4) \\[3ex] = x^2 + 4x + 2x + 8 \\[3ex] = x^2 + 6x + 8 \\[3ex] Compare:\;\; x^2 + mx + c \\[3ex] m = 6 \\[3ex] (x + 3)(x + 7) \\[3ex] = x^2 + 7x + 3x + 21 \\[3ex] = x^2 + 10x + 21 \\[3ex] Compare:\;\; x^2 + nx + d \\[3ex] n = 10 \\[3ex] mn \\[3ex] = 6(10) \\[3ex] = 60 $

(15.) Subtract the polynomials and simplify the result completely.
$(10x^5 + 11x^4 + 7x^3 - 1) - (-5x^4 - 4x^2 - 9x + 6)$

We can solve this question using at least two approaches.
Use any approach you prefer.

$ (10x^5 + 11x^4 + 7x^3 - 1) - (-5x^4 - 4x^2 - 9x + 6) \\[3ex] \underline{First\:\:Approach:\;\;Horizontal\;\;Method} \\[3ex] (10x^5 + 11x^4 + 7x^3 - 1) - (-5x^4 - 4x^2 - 9x + 6) \\[3ex] = 1(10x^5 + 11x^4 + 7x^3 - 1) - 1(-5x^4 - 4x^2 - 9x + 6) \\[3ex] Distributive\:\:Property \\[3ex] = 10x^5 + 11x^4 + 7x^3 - 1 + 5x^4 + 4x^2 + 9x - 6 \\[3ex] Collect\:\:like\:\:terms \\[3ex] = 10x^5 + 11x^4 + 5x^4 + 7x^3 + 4x^2 + 9x - 1 - 6 \\[3ex] = 10x^5 + 16x^4 + 7x^3 + 4x^2 + 9x - 7 \\[5ex] \underline{Second\:\:Approach:\;\;Vertical\;\;Method} \\[3ex] \begin{array}{c} 10x^5 + 11x^4 + 7x^3 + 0x^2 + 0x - 1 \\ -\hspace{16em} \\ 0x^5 - 5x^4 + 0x^3 - 4x^2 - 9x + 6 \\ \hline 10x^5 + 16x^4 + 7x^3 + 4x^2 + 9x - 7 \\ \hline \end{array} $

(16.) Divide $\dfrac{25x^7 + 10x^3}{-x^3}$

$ \dfrac{25x^7 + 10x^3}{-x^3} \\[5ex] = \dfrac{25x^7}{-x^3} + \dfrac{10x^3}{-x^3} \\[5ex] = -\dfrac{25x^7}{x^3} - \dfrac{10x^3}{x^3} \\[5ex] = -25x^{7 - 3} - 10x^{3 - 3} ...Law\;2...Exp \\[3ex] = -25x^4 - 10x^0 \\[3ex] = -25x^4 - 10(1) ...Law\;4...Exp \\[3ex] = -25x^4 - 10 \\[3ex] $ Student: Mr. C, could you not have just cancelled the x³?
Rather than applying Law 2 and then Law 4?
Teacher: Yes, you may choose to do it that way.

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