# Solved Examples: Factoring

As applicable, verify your answers with the Expressions and Equations Calculators

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 any 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.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE and Malta Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

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.

Factor all expressions
Indicate the factoring technique you use
If the expression cannot be factored; state so, and give reasons
Show all work
For all Quadratic Trinomials and the Likes that can be factored, check your solution
For all Quadratic Trinomials and the Likes in which the coefficient of the first term is $1$, you may use the method explained in *Step $3$* on Factor Quadratic Trinomials

(1.) $-13k^2 - 52$

$-13k^2 - 52 \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -13 \\[3ex] = -13(k^2 + 4)$
(2.) $-6a^5 + 12a^4 - 9a^3$

$-6a^5 + 12a^4 - 9a^3 \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -3a^3 \\[3ex] = -3a^3(2a^2 - 4a + 3)$
(3.) $9p^2 + 27p^3 - 18p^4$

$9p^2 + 27p^3 - 18p^4 \\[3ex] Arrange\: in\: descending\: order\: of\: exponents \\[3ex] -18p^4 + 27p^3 + 9p^2 \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -9p^2 \\[3ex] = -9p^2(2p^2 - 3p - 1)$
(4.) $(k - 5)2k + (k - 5)3$

$(k - 5)2k + (k - 5)3 \\[3ex] Arrange\: it\: well \\[3ex] 2k(k - 5) + 3(k - 5) \\[3ex] Factor\: by\: GCF \\[3ex] GCF = (k - 5) \\[3ex] = (k - 5)[2k + 3] \\[3ex] = (k - 5)(2k + 3)$
(5.) $3c(c^2 + 5) + 7(c^2 + 5)$

$3c(c^2 + 5) + 7(c^2 + 5) \\[3ex] Factor\: by\: GCF \\[3ex] GCF = c^2 + 5 \\[3ex] = (c^2 + 5)[3c + 7] \\[3ex] = (c^2 + 5)(3c + 7)$
(6.) $6g(g - 3) - 2(g - 3)$

$6g(g - 3) - 2(g - 3) \\[3ex] Factor\: by\: GCF \\[3ex] GCF = (g - 3) \\[3ex] = (g - 3)[6g - 2] \\[3ex] For (6g - 2); \\[3ex] Factor\: by\: GCF \\[3ex] GCF = 2 \\[3ex] 6g - 2 = 2(3g - 1) \\[3ex] = (g - 3)(2)(3g - 1) \\[3ex] = 2(g - 3)(3g - 1)$
(7.) $6a^3 - 9a^2 - 2a + 3$

$6a^3 - 9a^2 - 2a + 3 \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 6a^3 - 9a^2 = 3a^2(2a - 3) \\[3ex] -2a + 3 = -1(2a - 3) \\[3ex] = (2a - 3)(3a^2 - 1)$
(8.) $3y^2 - xy - 6y + 2x$

$3y^2 - xy - 6y + 2x \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 3y^2 - xy = y(3y - x) \\[3ex] -6y + 2x = -2(3y - x) \\[3ex] = (3y - x)(y - 2)$
(9.) $4d^3 + 6d^2 - 2d - 3$

$4d^3 + 6d^2 - 2d - 3 \\[3ex] Factor\: by\: Grouping \\[3ex] Group\: in\: twos \\[3ex] Use\: Vertical\: Method \\[3ex] 4d^3 + 6d^2 = 2d^2(2d + 3) \\[3ex] -2d - 3 = -1(2d + 3) \\[3ex] = (2d + 3)(2d^2 - 1)$
(10.) $-12p - 45d$

$-12p -45d \\[3ex] Factor\: by\: GCF \\[3ex] GCF = -3 \\[3ex] = -3(4p + 15d)$
(11.) $-k^2 + 13k - 42$

$-k^2 + 13k - 42 \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] = -1(k^2 - 13k + 42) \\[3ex] k^2 - 13k + 42 \\[3ex] Compare\: to\: ak^2 + bk + c \\[3ex] a = 1, b = -13, c = 42 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-13)^2 - 4(1)(42) \\[3ex] = 169 - 168 = 1 \\[3ex] 1 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] (k^2)(42) = 42k^2 \\[3ex] Factors\: are\: -6k, -7k \\[3ex] k^2 - 6k - 7k + 42 \\[3ex] k^2 - 6k = k(k - 6) \\[3ex] -7k + 42 = -7(k - 6) \\[3ex] (k - 6)(k -7) \\[3ex] = -1(k - 6)(k - 7) \\[3ex]$ Check your solution

$-1(k - 6)(k - 7) \\[3ex] = -1(k^2 - 7k - 6k + 42) \\[3ex] = -1(k^2 - 13k + 42) \\[3ex] = -k^2 + 13k - 42$
(12.) $x^2 - 144$

$x^2 - 144 \\[3ex] = x^2 - 12^2 \\[3ex] Factor\;\;by\;\;Difference\;\;of\;\;Two\;\;Squares \\[3ex] Compare\: to\: x^2 - y^2 \\[3ex] x = x \\[3ex] y = 12 \\[3ex] = (x + 12)(x - 12) \\[3ex] \underline{Check\;\;solution} \\[3ex] (x + 12)(x - 12) \\[3ex] = x^2 - 12x + 12x - 144 \\[3ex] = x^2 - 144$
(13.) $-3 + 20p^2 -11p$

$-3 + 20p^2 -11p \\[3ex] = 20p^2 - 11p - 3 \\[3ex] Compare\: to\: ap^2 + bp + c \\[3ex] a = 20, b = -11, c = -3 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-11)^2 - 4(20)(-3) \\[3ex] = 121 + 240 = 361 \\[3ex] 361 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] (20p^2)(-3) = -60p^2 \\[3ex] Factors\: are\: 4p, -15p \\[3ex] 20p^2 - 11p - 3 \\[3ex] = 20p^2 + 4p - 15p - 3 \\[3ex] 20p^2 + 4p = 4p(5p + 1) \\[3ex] -15p - 3 = -3(5p + 1) \\[3ex] (5p + 1)(4p - 3) \\[3ex]$ Check your solution

$(5p + 1)(4p - 3) \\[3ex] = 20p^2 - 15p + 4p - 3 \\[3ex] = 20p^2 - 11p -3 \\[3ex] = -3 + 20p^2 -11p$
(14.) $m^2 + 10m + 15$

$m^2 + 10m + 15 \\[3ex] a = 1, b = 10, c = 15 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 10^2 - 4(1)(15) \\[3ex] = 100 - 60 = 40 \\[3ex] 40 \:\:is\:\:not\:\:a\:\: Perfect\: Square \\[3ex] Trinomial\: cannot\: be\: factored \\[3ex] Trinomial\: is\: prime$
(15.) $q^2 + 12 + 13q$

$q^2 + 12 + 13q \\[3ex] = q^2 + 13q + 12 \\[3ex] Compare\: to\: aq^2 + bq + c \\[3ex] a = 1, b = 13, c = 12 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 13^2 - 4(1)(12) \\[3ex] = 169 - 48 = 125 \\[3ex] 125 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] (q^2)(12) = 12q^2 \\[3ex] Factors\: are\: q, 12q \\[3ex] q^2 + q + 12q + 12 \\[3ex] q^2 + q = q(q + 1) \\[3ex] 12q + 12 = 12(q + 1) \\[3ex] (q + 1)(q + 12) \\[3ex]$ OR use the method explained in *Step $3$* on "Factor Quadratic Trinomials"

$q^2 + 12 + 13q \\[3ex] (q + 1)(q + 12) \\[3ex]$ Check your solution

$(q + 1)(q + 12) \\[3ex] = q^2 + 12q + q + 12 \\[3ex] = q^2 + 13q + 12 \\[3ex] = q^2 + 12 + 13q$
(16.) $d^4 - 17d^3 + 60d^2$

$d^4 - 17d^3 + 60d^2 \\[3ex] Factor\:\:by\:\:GCF \\[3ex] GCF = d^2 \\[3ex] = d^2(d^2 - 17d + 60) \\[3ex]$ Let us deal with $d^2 - 17d + 60$
We shall include the $d^2$ in our final answer

$d^2 - 17d + 60 \\[3ex] Compare\: to\: ad^2 + bd + c \\[3ex] a = 1, b = -17, c = 60 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] Discriminant = (-17)^2 - 4(1)(60) \\[3ex] Discriminant = 289 - 240 = 49 \\[3ex] 49 \:\:is\:\:a\:\: Perfect\: Square \\[3ex] Factor\: Quadratic\: Trinomials \\[3ex] (d^2)(60) = 60d^2 \\[3ex] Factors\: are\: -5d, -12d \\[3ex] (d - 5)(d - 12) \\[3ex] \rightarrow d^4 - 17d^3 + 60d^2 \\[3ex] = d^2(d - 5)(d - 12) \\[3ex]$ Check your solution

$d^2(d - 5)(d - 12) \\[3ex] = d^2(d^2 - 12d - 5d + 60) \\[3ex] = d^2(d^2 - 17d + 60) \\[3ex] = d^4 - 17d^3 + 60d^2$
(17.) NZQA Simplify, as far as possible,

$\dfrac{4x^2 - 25}{2x^2 - x - 10} \\[5ex]$

$\underline{Numerator} \\[3ex] 4x^2 - 25 \\[3ex] = 2^2x^2 - 5^2 \\[3ex] = (2x)^2 - 5^2 \\[3ex] Factor\;\;by\;\;Difference\;\;of\;\;Two\;\;Squares \\[3ex] = (2x + 5)(2x - 5) \\[3ex] \underline{Denominator} \\[3ex] 2x^2 - x - 10 \\[3ex] (2x^2)(-10) = -20x^2 \\[3ex] Factors\;\;are\;\; 4x\;\;and\;\;-5x \\[3ex] Factor\;\;Quadratic\;\;Trinomial \\[3ex] = 2x^2 + 4x - 5x - 10 \\[3ex] = 2x(x + 2) -5(x + 2) \\[3ex] = (x + 2)(2x - 5) \\[3ex] \implies \\[3ex] \dfrac{4x^2 - 25}{2x^2 - x - 10} \\[5ex] = \dfrac{(2x + 5)(2x - 5)}{(x + 2)(2x - 5)} \\[5ex] = \dfrac{2x + 5}{x + 2}$
(18.) CSEC Factorize completely:

$(i)\;\; x^2 - 5x + 4 \\[3ex] (ii)\;\; m^2 - 4n^2 \\[3ex]$

$(i) \\[3ex] x^2 - 5x + 4 \\[3ex] Factor\;\;Quadratic\;\;Trinomial...Using\;\;Step\;\;3 \\[3ex] x^2 * 4 = 4x^2 \\[3ex] Factors\;\;are\;\;-1\;\;and\;\;-4 \\[3ex] = (x - 1)(x - 4) \\[3ex] \underline{Check\;\;solution} \\[3ex] (x - 1)(x - 4) \\[3ex] = x^2 - 4x - x + 4 \\[3ex] = x^2 - 5x + 4 \\[3ex] (ii) \\[3ex] m^2 - 4n^2 \\[3ex] = m^2 - 2^2n^2 \\[3ex] = m^2 - (2n)^2 \\[3ex] Factor\;\;by\;\;Difference\;\;of\;\;Two\;\;Squares \\[3ex] = (m + 2n)(m - 2n) \\[3ex] \underline{Check\;\;solution} \\[3ex] (m + 2n)(m - 2n) \\[3ex] = m^2 - 2mn + 2mn - (2n)^2 \\[3ex] = m^2 - 4n^2$
(19.)

(20.) KCSE Simplify $\dfrac{2x^2 - xy - 6y^2}{x^2 - 4xy + 4y^2}$

Factor Quadratic Trinomials in Two Variables

$\underline{Numerator} \\[3ex] 2x^2 - xy - 6y^2 \\[3ex] (2x^2)(-6y^2) = -12x^2y^2 \\[3ex] Factors = -4xy \;\;and\;\; 3xy \\[3ex] = 2x^2 - 4xy + 3xy - 6y^2 \\[3ex] = 2x(x - 2y) + 3y(x - 2y) \\[3ex] = (x - 2y)(2x + 3y) \\[3ex] \underline{Denominator} \\[3ex] x^2 - 4xy + 4y^2 \\[3ex] (x^2)(4y^2) = 4x^2y^2 \\[3ex] Factors = -2xy \;\;and\;\; -2xy \\[3ex] = x^2 - 2xy - 2xy + 4y^2 \\[3ex] = x(x - 2y) - 2y(x - 2y) \\[3ex] = (x - 2y)(x - 2y) \\[3ex] \implies \\[3ex] \dfrac{2x^2 - xy - 6y^2}{x^2 - 4xy + 4y^2} \\[5ex] = \dfrac{(x - 2y)(2x + 3y)}{(x - 2y)(x - 2y)} \\[5ex] = \dfrac{2x + 3y}{x - 2y}$

(21.)

(22.) $w^6 - 100$

$w^6 - 100 \\[3ex] = (w^3)^2 - 10^2 \\[3ex] Factor\;\;by\;\;Difference\;\;of\;\;Two\;\;Squares \\[3ex] Compare\: to\: x^2 - y^2 \\[3ex] x = w^3 \\[3ex] y = 10 \\[3ex] = (w^3 + 10)(w^3 - 10) \\[3ex] \underline{Check\;\;solution} \\[3ex] (w^3 + 10)(w^3 - 10) \\[3ex] = w^6 - 10w^3 + 10w^3 - 100 \\[3ex] = w^6 - 100$
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