Samuel Dominic Chukwuemeka (SamDom For Peace)

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Chukwuemeka



Solved Examples - Factoring

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 JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

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] Compare\: to\: x^2 - y^2 \\[3ex] x = x \\[3ex] y = 12 \\[3ex] = (x + 12)(x - 12) $
(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 $