Solved Examples on the Factoring of 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 GCSE Students
Questions on Factoring for both the Foundation Tier and Higher Tier (as applicable) are done per year (rather than separatly)

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.
Indicate the Factoring Technique(s) used for each question.
Show all work.
Check your solution for any question involving the factoring of quadratic trinomials. (Specifically for my Students unless specified otherwise.)
If the question involves the Factoring of Quadratic Trinomial, check your solution by multiplying the two binomials.

NOTES

$ \underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex] \underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex] (2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex] \underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex] (3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[5ex] $ Assume all expressions are polynomials

(1.) GCSE
(a.) Factorise fully 2x² + 6x
(b.) Factorise fully 144 − 4x²
(c.) Factorise 3x² + 11x − 20

(a.) is factored by GCF
(b.) is factored by GCF and Difference of Two Squares
(c.) is factored by Factor Quadratic Trinomial

$ (a.) \\[3ex] 2x^2 + 6x \\[3ex] GCF = 2x \\[3ex] \implies \\[3ex] 2x(x + 3) \\[5ex] (b.) \\[3ex] 144 - 4x^2 \\[3ex] GCF = 4 \\[3ex] \implies \\[3ex] 4(36 - x^2) \\[3ex] Keep\;\; 4\;\;for\;\;now \\[3ex] 36 - x^2 \\[3ex] 6^2 - x^2 \\[3ex] Compare\;\;to:\;\;x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 6 \\[3ex] y = x \\[3ex] \implies \\[3ex] (6 + x)(6 - x) \\[3ex] Bring\;\;down\;\;4 \\[3ex] 4(6 + x)(6 - x) \\[5ex] (c.) \\[3ex] 3x^2 + 11x - 20 \\[3ex] (3x^2)(-20) = -60x^2 \\[3ex] Factors\;\;are\;\;15x\;\;and\;\;-4x \\[3ex] \implies \\[3ex] 3x^2 + 15x - 4x - 20 \\[3ex] 3x(x + 5) - 4(x + 5) \\[3ex] (x + 5)(3x - 4) \\[3ex] \underline{Check} \\[3ex] (x + 5)(3x - 4) \\[3ex] 3x^2 - 4x + 15x - 20 \\[3ex] 3x^2 + 11x - 20 $

(2.) Factor the quadratic trinomials:

$ (a.)\;\; 7 - 8x + x^2 \\[3ex] (b.)\;\; -20 + j^2 + j \\[3ex] $

We have to arrange each of them in standard form.
Both questions are cases where the coefficient of the first term is unity.
So, we can just write it this way: (x ... ) and (x ... ) and include the two factors of the third term whose sum gives the second term.

$ (a.) \\[3ex] 7 - 8x + x^2 \\[3ex] Standard\;\;Form:\;\;x^2 - 8x + 7 \\[3ex] Factors\;\;are\;\;-1\;\;and\;\;-7 \\[3ex] \implies \\[3ex] (x - 1)(x - 7) \\[3ex] \underline{Check} \\[3ex] (x - 1)(x - 7) \\[3ex] F:\;\; x(x) = x^2 \\[3ex] O:\;\; x(-7) = -7x \\[3ex] I:\;\; -1(x) = -x \\[3ex] L:\;\; (-1)(-7) = 7 \\[3ex] \implies \\[3ex] x^2 - 7x - x + 7 \\[3ex] x^2 - 8x + 7 \\[3ex] (b.) \\[3ex] -20 + j^2 + j \\[3ex] Standard\;\;Form:\;\;j^2 + j - 20 \\[3ex] Factors\;\;are\;\;5\;\;and\;\;-4 \\[3ex] \implies \\[3ex] (j + 5)(j - 4) \\[3ej] \underline{Check} \\[3ej] (j + 5)(j - 4) \\[3ej] F:\;\; j(j) = j^2 \\[3ej] O:\;\; j(-4) = -4j \\[3ej] I:\;\; 5(j) = 5j \\[3ej] L:\;\; 5(-4) = -20 \\[3ej] \implies \\[3ej] j^2 - 4j + 5j - 20 \\[3ej] j^2 + j - 20 $

(3.) ACT Given that x² − 5x − 36 factors into 2 binomial factors with integer coefficients, which of the following binomials is 1 of those factors?

$ F.\;\; x - 12 \\[3ex] G.\;\; x - 9 \\[3ex] H.\;\; x - 4 \\[3ex] J.\;\; x + 6 \\[3ex] K.\;\; x + 12 \\[3ex] $

$ x^2 - 5x - 36 \\[3ex] x^2(-36) = -36x^2 \\[3ex] Factors\;\;are\;\;4x\;\;and\;\;-9x \\[3ex] \implies \\[3ex] (x + 4)(x - 9) \\[3ex] $ Answer = Option G.

(4.) JAMB Factorise 6x² − 14x − 12

$ A.\;\; 2(x + 3)(3x - 2) \\[3ex] B.\;\; 6(x - 2)(x + 1) \\[3ex] C.\;\; 2(x - 3)(3x + 2) \\[3ex] D.\;\; 6(x + 2)(x - 1) \\[3ex] E.\;\; (3x - 4)(2x + 3) \\[3ex] $

Factor by GCF then Factor Quadratic Trinomial

$ f(x) = 6x^2 - 14x - 12 \\[3ex] GCF = 2 \\[3ex] f(x) = 2(3x^2 - 7x - 6) \\[3ex] Keep\;\;2\;\;for\;\;now \\[3ex] 3x^2 - 7x - 6 \\[3ex] (3x^2)(-6) = -18x^2 \\[3ex] Factors\;\;are\;\;-9x\;\;and\;\;2x \\[3ex] \implies \\[3ex] 3x^2 - 9x + 2x - 6 \\[3ex] 3x(x - 3) + 2(x - 3) \\[3ex] (x - 3)(3x + 2) \\[3ex] Bring\;\;down\;\;2 \\[3ex] 2(x - 3)(3x + 2) \\[3ex] \therefore 6x^2 - 14x - 12 = 2(x - 3)(3x + 2) \\[3ex] \underline{Check} \\[3ex] 2(x - 3)(3x + 2) \\[3ex] 2(3x^2 + 2x - 9x - 6) \\[3ex] 2(3x^2 - 7x - 6) \\[3ex] 6x^2 - 14x - 12 $

(5.) NYSED The expression 36x² − 9 is equivalent to
(1) (6x − 3)² (3) (6x + 3)(6x − 3)
(2) (18x − 4.5)² (4) (18x + 4.5)(18x − 4.5)

$ 36x^2 - 9 \\[3ex] 6^2x^2 - 3^2 \\[3ex] (6x)^2 - 3^2 \\[3ex] Compare:\;\; x^2 - y^2 \\[3ex] x = 6x \\[3ex] y = 3 \\[3ex] \implies \\[3ex] (6x + 3)(6x - 3) \\[3ex] \underline{Check} \\[3ex] (6x + 3)(6x - 3) \\[3ex] 36x^2 - 18x + 18x - 9 \\[3ex] 36x^2 - 9 $

(6.) KCSE Simplify $(4 + 2y)^2 - (2y - 4)^2$

There are at least two ways to solve this question

$ \underline{First\;\;Appraoch:\;\;Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (4 + 2y)^2 - (2y - 4)^2 \\[3ex] Compare\;\;to:\;\; x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 4 + 2y \\[3ex] y = 2y - 4 \\[3ex] \implies \\[3ex] [(4 + 2y) + (2y - 4)][(4 + 2y) - (2y - 4)] \\[3ex] (4 + 2y + 2y - 4)(4 + 2y - 2y + 4) \\[3ex] (4y)(80) \\[3ex] 32y \\[3ex] \underline{Second\;\;Approach:\;\;Expand\;\;and\;\;Simplify} \\[3ex] (4 + 2y)^2 - (2y - 4)^2 \\[3ex] [(4 + 2y)(4 + 2y)] - [(2y - 4)(2y - 4)] \\[3ex] (16 + 8y + 8y + 4y^2) - (4y^2 - 8y - 8y + 16) \\[3ex] (16 + 16y + 4y^2) - (4y^2 - 16y + 16) \\[3ex] 16 + 16y + 4y^2 - 4y^2 + 16y - 16 \\[3ex] 32y $

(7.) JAMB Factorize completely $81a^4 - 16b^4$

$ A.\;\; (3a + 2b)(2a - 3b)(9a^2 + 4b^2) \\[3ex] B.\;\; (3a - 2b)(2a - 3b)(4a^2 - 9b^2) \\[3ex] C.\;\; (3a - 2b)(3a + 2b)(9a^2 + 4b^2) \\[3ex] D.\;\; (3a - 2b)(2a - 3b)(9a^2 + 4b^2) \\[3ex] E.\;\; (3a - 2b)(2a - 3b)(9a^2 - 4b^2) \\[3ex] $

Difference of Two Squares

$ 81a^4 - 16b^4 \\[3ex] 9^2(a^2)^2 - 4^2(b^2)^2 \\[3ex] (9a^2)^2 - (4b^2)^2 \\[3ex] Compare\;\;to:\;\; x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 9a^2 \\[3ex] y = 4b^2 \\[3ex] \implies \\[3ex] (9a^2 + 4b^2)(9a^2 - 4b^2) \\[3ex] Keep:\;\; 9a^2 + 4b^2 \\[3ex] Factor:\;\; 9a^2 - 4b^2 \\[3ex] 3^2a^2 - 2^2b^2 \\[3ex] (3a)^2 - (2b)^2 \\[3ex] Compare\;\;again \\[3ex] x = 3a \\[3ex] y = 2b \\[3ex] \implies \\[3ex] (3a + 2b)(3a - 2b) \\[3ex] \therefore 81a^4 - 16b^4 = (9a^2 + 4b^2)(3a + 2b)(3a - 2b) \\[3ex] = (3a - 2b)(3a + 2b)(9a^2 + 4b^2) $

(8.) GCSE Here is an identity. $$x^2 - y^2 \equiv (x + y)(x - y)$$ (a) Use the identity to work out the value of 193² − 7²
You must show your working.

(b) Factorise 100a² - 81b²

Difference of Two Squares

$ (a) \\[3ex] 193^2 - 7^2 \\[3ex] Compare\;\;to:\;\;x^2 - y^2 \equiv (x + y)(x - y) \\[3ex] x = 193 \\[3ex] y = 7 \\[3ex] \implies \\[3ex] (193 + 7)(193 - 7) \\[3ex] 200(186) \\[3ex] 37200 \\[3ex] (b) \\[3ex] 100a^2 - 81b^2 \\[3ex] 10^2a^2 - 9^2b^2 \\[3ex] (10a)^2 - (9b)^2 \\[3ex] Compare\;\;to:\;\;x^2 - y^2 \equiv (x + y)(x - y) \\[3ex] x = 10a \\[3ex] y = 9b \\[3ex] \implies \\[3ex] (10a + 9b)(10a - 9b) $

(9.) ACT One root of the quadratic polynomial $ax^2 + 13x - 6$ is equal to −3.
Which of the following binomials is a factor of $ax^2 + 13x - 6$?

$ F.\;\; x - \dfrac{2}{5} \\[5ex] G.\;\; x + \dfrac{2}{5} \\[5ex] H.\;\; x - 3 \\[3ex] J.\;\; x - 5 \\[3ex] K.\;\; x + 5 \\[3ex] $

$ ax^2 + 13x - 6 \\[3ex] Root:\;\; x = -3 \\[3ex] \implies \\[3ex] a(-3)^2 + 13(-3) - 6 = 0 \\[3ex] 9a - 39 - 6 = 0 \\[3ex] 9a = 39 + 6 \\[3ex] 9a = 45 \\[3ex] a = \dfrac{45}{9} \\[5ex] a = 5 \\[3ex] \implies \\[3ex] Expression:\;\; 5x^2 + 13x - 6 \\[3ex] Factor\;\;the\;\;expression \\[3ex] (5x^2)(-6) = -30x^2 \\[3ex] Factors\;\;are\;\; 15x\;\;and\;\;-2x \\[3ex] \implies \\[3ex] 5x^2 + 15x - 2x - 6 \\[3ex] 5x(x + 3) - 2(x + 3) \\[3ex] (x + 3)(5x - 2) \\[3ex] Factors\;\;= (x + 3) \;\;and\;\;(5x - 2) \\[3ex] Zeros: \\[3ex] x + 3 = 0 \\[3ex] x = -3...given \\[3ex] 5x - 2 = 0 \\[3ex] 5x = 2 \\[3ex] x = \dfrac{2}{5} \\[5ex] Zero\;\;of\;\;x = \dfrac{2}{5} \implies Factor\;\;of\;\; x - \dfrac{2}{5} \\[5ex] \therefore One\;\;Factor = x - \dfrac{2}{5} $

(10.) JAMB Factorize completely $8a + 125ax^3$

$ A.\;\; (2a + 5x^2)(4 + 26ax) \\[3ex] B.\;\; a(2 + 5x)(4 - 10x + 25x^2) \\[3ex] C.\;\; (2a + 5x)(4 - 10ax + 25x^2) \\[3ex] D.\;\; a(2 + 5x)(4 + 10ax + 25x^2) \\[3ex] $

GCF then Sum of Two Cubes

$ 8a + 125ax^3 \\[3ex] a(8 + 125x^3) \\[3ex] Keep:\;\;a \\[3ex] Factor:\;\; 8 + 125x^3 \\[3ex] 8 + 125x^3 \\[3ex] 2^3 + 5^3x^3 \\[3ex] 2^3 + (5x)^3 \\[3ex] Compare\;\;to:\;\;x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[3ex] x = 2 \\[3ex] y = 5x \\[3ex] \implies \\[3ex] (2 + 5x)[2^2 - 2(5x) + (5x)^2] \\[3ex] (2 + 5x)(4 - 10x + 25x^2) \\[3ex] Bring\;\;back:\;\;a \\[3ex] = a(2 + 5x)(4 - 10x + 25x^2) $

(11.) JAMB Find a factor which is common to all three binomial expressions $$4a^2 - 9b^2,\;\;\;8a^3 + 27b^3,\;\;\;(4a + 6b)^2$$ $ A.\;\; 4a + 6b \\[3ex] B.\;\; 4a - 6b \\[3ex] C.\;\; 2a + 3b \\[3ex] D.\;\; 2a - 3b \\[3ex] E.\;\; None \\[3ex] $

For:
4a² - 9b², we use Difference of Two Sqaures
8a³ + 27b³, we use Sum of Two Cubes
(4a + 6b)², we use Factoring by GCF

$ \underline{1st\;\;Binomial\;\;Expression} \\[3ex] 4a^2 - 9b^2 \\[3ex] 2^2a^2 - 3^2b^2 \\[3ex] (2a)^2 - (3b)^2 \\[3ex] Compare\;\;to:\;\;x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 2a \\[3ex] y = 3b \\[3ex] \implies \\[3ex] (2a + 3b)(2a - 3b) \\[5ex] \underline{2nd\;\;Binomial\;\;Expression} \\[3ex] 8a^3 + 27b^3 \\[3ex] 2^3a^3 + 3^3b^3 \\[3ex] (2a)^3 + (3b)^3 \\[3ex] Compare\;\;to:\;\;x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[3ex] x = 2a \\[3ex] y = 3b \\[3ex] \implies \\[3ex] (2a + 3b)[(2a)^2 - 2a(3b) + (3b)^2] \\[3ex] (2a + 3b)(4a^2 - 6ab + 9b^2) \\[5ex] \underline{3rd\;\;\;Binomial\;\;Expression} \\[3ex] (4a + 6b)^2 \\[3ex] [2(2a + 3b)]^2 \\[3ex] 2^2 * (2a + 3b)^2 \\[3ex] 4(2a + 3b)(2a + 3b) \\[5ex] \underline{All\;\;Three:\;\;Factor\;\;by\;\;GCF} \\[3ex] 4a^2 - 9b^2 = (2a + 3b)(2a - 3b) \\[3ex] 8a^3 + 27b^3 = (2a + 3b)(4a^2 - 6ab + 9b^2) \\[3ex] (4a + 6b)^2 = 4(2a + 3b)(2a + 3b) \\[3ex] GCF = 2a + 3b $

(12.) ACT Given that x = −2 is a solution to $x^2 + bx - 6 = 0$, which of the following polynomials is a factor of $x^2 + bx - 6$?

$ F.\;\; x - 3 \\[3ex] G.\;\; x - 2 \\[3ex] H.\;\; x - 1 \\[3ex] J.\;\; x + 1 \\[3ex] K.\;\; x + 3 \\[3ex] $

$ x^2 + bx - 6 = 0 \\[3ex] x = -2 \;\;is\;\;a\;\;solution \implies \\[3ex] (-2)^2 + b(-2) - 6 = 0 \\[3ex] 4 -2b - 6 = 0 \\[3ex] -2b - 2 = 0 \\[3ex] -2b = 2 \\[3ex] b = -\dfrac{2}{2} \\[5ex] b = -1 \\[3ex] \implies \\[3ex] Equation:\;\;x^2 - x - 6 = 0 \\[3ex] Expression:\;\; x^2 - x - 6 \\[3ex] (x^2)(-6) = -6x^2 \\[3ex] Factors\;\;are\;\;2x\;\;and\;\;-3x \\[3ex] \implies \\[3ex] (x + 2)(x - 3) \\[3ex] $ Answer = Option F.

(13.) For each of these quadratic trinomials:
(a.) Determine the discriminant
(b.) State whether it can be factored or not.
(c.) If it can be factored, factor it.
(d.) If you factor it, check your solution.

$ (i.)\;\; p^2 + 3p + 12 \\[3ex] (ii.)\;\; -4 - 47x + 12x^2 \\[3ex] (iii.)\;\; 20x^2 + 11x - 3 \\[3ex] (iv.)\;\; -3p^2 - p + 4 \\[3ex] (v.)\;\; 35 - 2x - x^2 \\[3ex] $

Factor Quadratic Trinomial

$ (i.) \\[3ex] p^2 + 3p + 12 \\[3ex] (a.) \\[3ex] a = 1 \\[3ex] b = 3 \\[3ex] c = 12 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 3^2 - 4(1)(12) \\[3ex] = 9 - 48 \\[3ex] = -39 \\[3ex] (b.) \\[3ex] The\;\;trinomial\;\;cannot\;\;be\;\;factored\;\;because\;\;the\;\;discriminant\;\;is\;\;not\;\;a\;\;perfect\;\;square \\[5ex] (ii.) \\[3ex] -4 - 47x + 12x^2 \\[3ex] 12x^2 - 47x - 4 \\[3ex] a = 12 \\[3ex] b = -47 \\[3ex] c = -4 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-47)^2 - 4(12)(-4) \\[3ex] = 2209 - (-192) \\[3ex] = 2209 + 192 \\[3ex] = 2401 = 49^2 \\[3ex] (b.) \\[3ex] Because\;\;the\;\;discriminant\;\;is\;\;a\;\;perfect\;\;square,\;\;the\;\;quadratic\;\;trinomial\;\;can\;\;be\;\;factored \\[3ex] (c.) \\[3ex] 12x^2 - 47x - 4 \\[3ex] 12x^2(-4) = -48x^2 \\[3ex] Factors = -48x + x \\[3ex] \implies \\[3ex] 12x^2 - 48x + x - 4 \\[3ex] 12x(x - 4) + 1(x - 4) \\[3ex] (x - 4)(12x + 1) \\[3ex] (d.) \\[3ex] \underline{Check} \\[3ex] (x - 4)(12x + 1) \\[3ex] 12x^2 + x - 48x - 4 \\[3ex] 12x^2 - 47x - 4 \\[3ex] -4 - 47x + 12x^2 \\[5ex] (iii.) \\[3ex] 20x^2 + 11x - 3 \\[3ex] a = 20 \\[3ex] b = 11 \\[3ex] c = -3 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = 11^2 - 4(20)(-3) \\[3ex] = 121 - (-240) \\[3ex] = 121 + 240 \\[3ex] = 361 = 19^2 \\[3ex] (b.) \\[3ex] Because\;\;the\;\;discriminant\;\;is\;\;a\;\;perfect\;\;square,\;\;the\;\;quadratic\;\;trinomial\;\;can\;\;be\;\;factored \\[3ex] (c.) \\[3ex] 20x^2 + 11x - 3 \\[3ex] 20x^2(-3) = -60x^2 \\[3ex] Factors = -4x + 15x \\[3ex] \implies \\[3ex] 20x^2 - 4x + 15x - 3 \\[3ex] 4x(5x - 1) + 3(5x - 1) \\[3ex] (5x - 1)(4x + 3) \\[3ex] (d.) \\[3ex] \underline{Check} \\[3ex] (5x - 1)(4x + 3) \\[3ex] 20x^2 + 15x - 4x - 3 \\[3ex] 20x^2 + 11x - 3 \\[5ex] (iv.) \\[3ex] -3p^2 - p + 4 \\[3ex] a = -3 \\[3ex] b = -1 \\[3ex] c = 4 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-1)^2 - 4(-3)(4) \\[3ex] = 1 - (-48) \\[3ex] = 1 + 48 \\[3ex] = 49 = 7^2 \\[3ex] (b.) \\[3ex] Because\;\;the\;\;discriminant\;\;is\;\;a\;\;perfect\;\;square,\;\;the\;\;quadratic\;\;trinomial\;\;can\;\;be\;\;factored \\[3ex] (c.) \\[3ex] -3p^2 - p + 4 \\[3ex] -1(3p^2 + p - 4) \\[3ex] Keep\;\;-1\;\;for\;\;now \\[3ex] For:\;\; 3p^2 + p - 4 \\[3ex] 3p^2(-4) = -12p^2 \\[3ex] Factors = -3p + 4p \\[3ex] \implies \\[3ex] 3p^2 - 3p + 4p - 4 \\[3ex] 3p(p - 1) + 4(p - 1) \\[3ex] (p - 1)(3p + 4) \\[3ex] Bring\;\;down\;\;-1 \\[3ex] -1(p - 1)(3p + 4) \\[3ex] (d.) \\[3ex] \underline{Check} \\[3ex] -1(p - 1)(3p + 4) \\[3ex] -1(3p^2 + 4p - 3p - 4) \\[3ex] -1(3p^2 + p - 4) \\[5ex] -3p^2 - p + 4 \\[5ex] (v.) \\[3ex] 35 - 2x - x^2 \\[3ex] -x^2 - 2x + 35 \\[3ex] a = -1 \\[3ex] b = -2 \\[3ex] c = 35 \\[3ex] Discriminant = b^2 - 4ac \\[3ex] = (-2)^2 - 4(-1)(35) \\[3ex] = 4 - (-140) \\[3ex] = 4 + 140 \\[3ex] = 144 = 12^2 \\[3ex] (b.) \\[3ex] Because\;\;the\;\;discriminant\;\;is\;\;a\;\;perfect\;\;square,\;\;the\;\;quadratic\;\;trinomial\;\;can\;\;be\;\;factored \\[3ex] (c.) \\[3ex] -x^2 - 2x + 35 \\[3ex] -1(x^2 + 2x - 35) \\[3ex] Keep\;\;-1\;\;for\;\;now \\[3ex] For:\;\; x^2 + 2x - 35 \\[3ex] x^2(-35) = -35x^2 \\[3ex] Factors = 7x - 5x \\[3ex] \implies \\[3ex] x^2 + 7x - 5x - 35 \\[3ex] x(x + 7) - 5(x + 7) \\[3ex] (x + 7)(x - 5) \\[3ex] Bring\;\;down\;\;-1 \\[3ex] -1(x + 7)(x - 5) \\[3ex] (d.) \\[3ex] \underline{Check} \\[3ex] -1(x + 7)(x - 5) \\[3ex] -1(x^2 - 5x + 7x - 35) \\[3ex] -1(x^2 + 2x - 35) \\[3ex] -x^2 - 2x + 35 \\[3ex] 35 - 2x - x^2 $

(14.) JAMB Factorize x² + 2a + ax + 2x

$ A.\;\; (x + 2a)(x + 1) \\[3ex] B.\;\; (x + 2a)(x - 1) \\[3ex] C.\;\; (x^2 - 1)(x - a) \\[3ex] D.\;\; (x + 2)(x + a) \\[3ex] $

Factor by Grouping

$ x^2 + 2a + ax + 2x \\[3ex] x^2 + ax + 2x + 2a \\[3ex] x(x + a) + 2(x + a) \\[3ex] (x + a)(x + 2) \\[3ex] (x + 2)(x + a) $

(15.) JAMB The factors of $9 - (x^2 - 3x - 1)^2$ are

$ A.\;\; -(x - 4)(x + 1)(x - 1)(x - 2) \\[3ex] B.\;\; (x - 4)(x - 1)(x - 1)(x + 2) \\[3ex] C.\;\; -(x - 2)(x + 1)(x + 2)(x + 4) \\[3ex] D.\;\; (x - 4)(x - 3)(x - 2)(x + 1) \\[3ex] E.\;\; (x - 2)(x + 2)(x - 1)(x + 1) \\[3ex] $

Difference of Two Squares and Factor Quadratic Trinomial

$ 9 - (x^2 - 3x - 1)^2 \\[3ex] 3^2 - (x^2 - 3x - 1)^2 \\[3ex] Compare\;\;to:\;\; x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 3 \\[3ex] y = x^2 - 3x - 1 \\[3ex] \implies \\[3ex] [3 + (x^2 - 3x - 1)][3 - (x^2 - 3x - 1)] \\[3ex] (3 + x^2 - 3x - 1)(3 - x^2 + 3x + 1) \\[3ex] (x^2 - 3x + 2)(-x^2 + 3x + 4) \\[3ex] (-x^2 + 3x + 4)(x^2 - 3x + 2) \\[3ex] Let\;\;us\;\;factor\;\;one\;\;at\;\;a\;\;time \\[3ex] First:\;\; -x^2 + 3x + 4 \\[3ex] -(x^2 - 3x - 4) \\[3ex] Keep\;\;-1 \\[3ex] x^2 - 3x - 4 \\[3ex] (x^2)(-4) = -4x^2 \\[3ex] Factors\;\;are\;\;x\;\;and\;\;-4x \\[3ex] \implies \\[3ex] x^2 + x - 4x - 4 \\[3ex] x(x + 1) -4(x + 1) \\[3ex] (x + 1)(x - 4) \\[3ex] Bring\;\;down\;\; -1 \\[3ex] -1(x + 1)(x - 4) \\[3ex] Second:\;\; x^2 - 3x + 2 \\[3ex] (x^2)(2) = 2x^2 \\[3ex] Factors\;\;are\;\;-x\;\;and\;\;-2x \\[3ex] \implies \\[3ex] x^2 - x - 2x + 2 \\[3ex] x(x - 1) - 2(x - 1) \\[3ex] (x - 1)(x - 2) \\[3ex] Combining\;\;everything\;\;together \\[3ex] 9 - (x^2 - 3x - 1)^2 \\[3ex] = -1(x + 1)(x - 4)(x - 1)(x - 2) \\[3ex] = -1(x - 4)(x + 1)(x - 1)(x - 2) $

(16.) ACT Which of the following binomials is a factor of $3x^2 + 11x - 4$

$ A.\;\; x - 4 \\[3ex] B.\;\; x - 2 \\[3ex] C.\;\; x - 1 \\[3ex] D.\;\; x + 2 \\[3ex] E.\;\; x + 4 \\[3ex] $

Factor Quadratic Trinomial

$ 3x^2 + 11x - 4 \\[3ex] 3x^2(-4) = -12x^2 \\[3ex] Factors = 12x - x \\[3ex] \implies \\[3ex] 3x^2 + 12x - x - 4 \\[3ex] 3x(x + 4) -1(x + 4) \\[3ex] (x + 4)(3x - 1) $

(18.) GCSE
(a) Simplify fully 3a² + 7a + 3 − a² + 8a − 4
(b) Factorise fully 24y² − 20y

(a) part can be solved by Simplifying
(b) part can be solved by Factor by GCF

$ (a) \\[3ex] 3a^2 + 7a + 3 - a^2 + 8a - 4 \\[3ex] 2a^2 + 15a - 1 \\[3ex] Cannot\;\;be\;\;simplified\;\;further \\[3ex] (b) \\[3ex] 24y^2 - 20y \\[3ex] 4y(6y - 5) $

(19.) Factor the trinomials completely.
Check your solutions.

$ (a.)\;\; 3x^2 + 30x - 72 \\[3ex] (b.)\;\; 4n^2 - 19n + 21 \\[3ex] (c.)\;\; p^2 + 2pq - 63q^2 \\[3ex] (d.)\;\; 2x^2 - 17xy + 30y^2 \\[3ex] (e.)\;\; x^2 - 14wx + 49w^2 \\[3ex] $

$ (a.) \\[3ex] 3x^2 + 30x - 72 \\[3ex] 3(x^2 + 10x - 24) \\[3ex] Keep\;\;3\;\;for\;\;now \\[3ex] For:\;\; x^2 + 10x - 24 \\[3ex] Factors = -2x + 12x \\[3ex] \implies \\[3ex] (x - 2)(x + 12) \\[3ex] Bring\;\;down\;\;3 \\[3ex] \implies \\[3ex] 3(x - 2)(x + 12) \\[3ex] \underline{Check} \\[3ex] 3(x - 2)(x + 12) \\[3ex] 3(x^2 + 12x - 2x - 24) \\[3ex] 3(x^2 + 10x - 24) \\[3ex] 3x^2 + 30x - 72 \\[5ex] (b.) \\[3ex] 4n^2 - 19n + 21 \\[3ex] 4n^2(21) = 84n^2 \\[3ex] Factors = -12n - 7n \\[3ex] \implies \\[3ex] 4n^2 - 12n - 7n + 21 \\[3ex] 4n(n - 3) - 7(n - 3) \\[3ex] (n - 3)(4n - 7) \\[3ex] \underline{Check} \\[3ex] (n - 3)(4n - 7) \\[3ex] 4n^2 - 7n - 12n + 21 \\[3ex] 4n^2 - 19n + 21 \\[5ex] (c.) \\[3ex] p^2 + 2pq - 63q^2 \\[3ex] p^2(-63q^2) = -63p^2q^2 \\[3ex] Factors = 9pq - 7pq \\[3ex] \implies \\[3ex] p^2 + 9pq - 7pq - 63q^2 \\[3ex] p(p + 9q) - 7q(p + 9q) \\[3ex] (p + 9q)(p - 7q) \\[3ex] \underline{Check} \\[3ex] (p + 9q)(p - 7q) \\[3ex] p^2 - 7pq + 9pq - 63q^2 \\[3ex] p^2 + 2pq - 63q^2 \\[5ex] (d.) \\[3ex] 2x^2 - 17xy + 30y^2 \\[3ex] 2x^2(30y^2) = 60x^2y^2 \\[3ex] Factors = -12xy - 5xy \\[3ex] \implies \\[3ex] 2x^2 - 12xy - 5xy + 30y^2 \\[3ex] 2x(x - 6y) - 5y(x - 6y) \\[3ex] (x - 6y)(2x - 5y) \\[3ex] \underline{Check} \\[3ex] (x - 6y)(2x - 5y) \\[3ex] 2x^2 - 5xy - 12xy + 30y^2 \\[3ex] 2x^2 - 17xy + 30y^2 \\[5ex] (e.) \\[3ex] x^2 - 14wx + 49w^2 \\[3ex] x^2(49w^2) = 49w^2x^2 \\[3ex] Factors = -7wx - 7wx \\[3ex] \implies \\[3ex] x^2 - 7wx - 7wx + 49w^2 \\[3ex] x(x - 7w) -7w(x - 7w) \\[3ex] (x - 7w)(x - 7w) \\[3ex] (x - 7w)^2 \\[3ex] \underline{Check} \\[3ex] (x - 7w)^2 \\[3ex] (x - 7w)(x - 7w) \\[3ex] x^2 - 7wx - 7wx + 49w^2 \\[3ex] x^2 - 14wx + 49w^2 $

(20.) JAMB Factorize $abx^2 + 8y - 4bx - 2axy$

$ A.\;\; (ax - 4)(bx - 2y) \\[3ex] B.\;\; (ax + b)(x - 8y) \\[3ex] C.\;\; (ax - 2y)(by - 4) \\[3ex] D.\;\; (abx - 4)(x - 2y) \\[3ex] E.\;\; (bx - 4)(ax - 2y) \\[3ex] $

Factor by Grouping then Factor by GCF

$ abx^2 + 8y - 4bx - 2axy \\[3ex] abx^2 - 4bx + 8y - 2axy \\[3ex] bx(ax - 4) + 2y(4 - ax) \\[3ex] $ Because $(ax - 4)$ is not the same as $(4 - ax)$, we need to multiply either one by -1 to get the other
Then, we also have to multiply the coefficient of that binomial by -1 to nullify the effect
Looking at the options, it is do this to the second part

$ -1 * -1 * 2y(4 - ax) \\[3ex] -1 * 2y * -1 * (4 - ax) \\[3ex] -1(4 - ax) = -4 + ax = ax - 4 \\[3ex] -1 * 2y = -2y \\[3ex] \implies \\[3ex] bx(ax - 4) - 2y(ax - 4) \\[3ex] = (ax - 4)(bx - 2y) \\[3ex] \underline{Check} \\[3ex] (ax - 4)(bx - 2y) \\[3ex] abx^2 - 2axy - 4bx + 8y \\[3ex] abx^2 + 8y - 4bx - 2axy $

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

(23.) JAMB Factorize $(4a + 3)^2 - (3a - 2)^2$

$ A.\;\; (a + 1)(a + 5) \\[3ex] B.\;\; (a - 5)(7a - 1) \\[3ex] C.\;\; (a + 5)(7a + 1) \\[3ex] D.\;\; a(7a + 1) \\[3ex] $

There are at least two ways to solve this question

$ \underline{First\;\;Appraoch:\;\;Difference\;\;of\;\;Two\;\;Squares} \\[3ex] (4a + 3)^2 - (3a - 2)^2 \\[3ex] Compare\;\;to:\;\; x^2 - y^2 = (x + y)(x - y) \\[3ex] x = 4a + 3 \\[3ex] y = 3a - 2 \\[3ex] \implies \\[3ex] [(4a + 3) + (3a - 2)][(4a + 3) - (3a - 2)] \\[3ex] (4a + 3 + 3a - 2)(4a + 3 - 3a + 2) \\[3ex] (7a + 1)(a + 5) \\[3ex] (a + 5)(7a + 1) \\[3ex] \underline{Second\;\;Approach:\;\;Expand\;\;and\;\;Factor\;\;Quadratic\;\;Trinomial} \\[3ex] (4a + 3)^2 - (3a - 2)^2 \\[3ex] [(4a + 3)(4a + 3)] - [(3a - 2)(3a - 2)] \\[3ex] (16a^2 + 12a + 12a + 9) - (9a^2 - 6a - 6a + 4) \\[3ex] (16a^2 + 24a + 9) - (9a^2 - 12a + 4) \\[3ex] 16a^2 + 24a + 9 - 9a^2 + 12a - 4 \\[3ex] 7a^2 + 36a + 5 \\[3ex] (7a^2)(5) = 35a^2 \\[3ex] Factors\;\;are\;\; 35a\;\;and\;\;a \\[3ex] \implies \\[3ex] 7a^2 + 35a + a + 5 \\[3ex] 7a(a + 5) + 1(a + 5) \\[3ex] (a + 5)(7a + 1) $

(26.) Factorize $(k - 5)2k + (k - 5)3$

Factor by GCF

$ (k - 5)2k + (k - 5)3 \\[3ex] 2k(k - 5) + 3(k - 5) \\[3ex] GCF = k - 5 \\[3ex] \implies \\[3ex] (k - 5)(2k + 3) $

(28.) Factor completely: $6g(g - 3) - 2(g - 3)$

Factor by Grouping $ 6g(g - 3) - 2(g - 3) \\[3ex] GCF = g - 3 \\[3ex] (g - 3)(6g - 2) \\[3ex] GCF\;\;of\;\;6g - 2 = 2 \\[3ex] (g - 3) * 2(3g - 1) \\[3ex] 2(g - 3)(3g - 1) $

(30.) Factorize $6a^3 - 9a^2 - 2a + 3$

Factor by Grouping

$ 6a^3 - 9a^2 - 2a + 3 \\[3ex] 6a^3 - 9a^2 = 3a^2(2a - 3) \\[3ex] -2a + 3 = -1(2a - 3) \\[3ex] \implies \\[3ex] (2a - 3)(3a^2 - 1) $

(32.) ACT Which of the following expressions is a factor of $x^3 - 64$?

$ A.\;\; x - 4 \\[3ex] B.\;\; x + 4 \\[3ex] C.\;\; x + 64 \\[3ex] D.\;\; x^2 + 16 \\[3ex] E.\;\; x^2 - 4x + 16 \\[3ex] $

Difference of Two Cubes

$ x^3 - 64 \\[3ex] x^3 - 4^3 \\[3ex] Compare\;\;to:\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[3ex] x = x \\[3ex] y = 4 \\[3ex] \implies \\[3ex] (x - 4)[x^2 + x(4) + 4^2] \\[3ex] (x - 4)(x^2 + 4x + 16) \\[3ex] $ Correct Option = Option A.

$ \underline{Check} \\[3ex] (x - 4)(x^2 + 4x + 16) \\[3ex] x^3 + 4x^2 + 16x - 4x^2 - 16x - 64 \\[3ex] x^3 - 64 $

(34.) NYSED Factor completely: 3y² − 12y − 288

$ 3y^2 - 12y - 288 \\[3ex] 3(y^2 - 4y - 96) \\[3ex] Keep\;\;3\;\;for\;\;now \\[3ex] y^2 - 4y - 96 \\[3ex] Factors = 8y - 12y \\[3ex] \implies \\[3ex] (y + 8)(y - 12) \\[3ex] Bring\;\;down\;\; 3 \\[3ex] 3(y + 8)(y - 12) \\[3ex] \underline{Check} \\[3ex] 3(y + 8)(y - 12) \\[3ex] 3(y^2 - 12y + 8y - 96) \\[3ex] 3(y^2 - 4y - 96) \\[3ex] 3y^2 - 12y - 288 $

(39.) JAMB Factorize $6^{2x + 1} + 7(6^x) - 5$

$ A.\;\; [3(6^x) - 5][2(6^x) + 1] \\[3ex] B.\;\; [3(6^x) + 5][2(6^x) - 1] \\[3ex] C.\;\; [(6^x) - 5][3(6^x) + 1] \\[3ex] D.\;\; [2(6^x) + 5][3(6^x) - 1] \\[3ex] $

Prior Knowledge of the Laws of Exponents and Factor Quadratic Trinomial

$ 6^{2x + 1} + 7(6^x) - 5 \\[3ex] 6^{2x} * 6 + 7(6^x) - 5 \\[3ex] 6 * 6^{2x} + 7(6^x) - 5 \\[3ex] 6(6^x)^2 + 7(6^x) - 5 \\[3ex] Let\;\; 6^x = p \\[3ex] Substitute\;\;6^x = p \\[3ex] \implies \\[3ex] 6p^2 + 7p - 5 \\[3ex] (6p^2)(-5) = -30p^2 \\[3ex] Factors = -3p\;\;and\;\;10p \\[3ex] \implies \\[3ex] 6p^2 - 3p + 10p - 5 \\[3ex] 3p(2p - 1) + 5(2p - 1) \\[3ex] (2p - 1)(3p + 5) \\[3ex] Resubstitute\;\;p \\[3ex] (2 * 6^x - 1)(3 * 6^x + 5) \\[3ex] [3(6^x) + 5][2(6^x) - 1] $

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