Factoring Polynomials: Difficult Problems with Solutions

Solution:
[tex]2x^{2} + 7x + 5 =[/tex]
[tex]2x^{2} + 2x + 5x + 5 =[/tex]
[tex]2x \times x + 2x \times 1 + 5 \times x + 5 \times 1 =[/tex]
[tex]2x(x + 1) + 5(x + 1) = (x + 1)(2x + 5)[/tex]

Solution:
[tex]3x^{2} - 4x - 7 =[/tex]
[tex]3x^{2} - 7x + 3x - 7 =[/tex]
[tex]x \times 3x + x \times (-7) + 1 \times 3x + 1 \times (-7) =[/tex]
[tex]x(3x - 7) + 1(3x - 7) = (3x - 7)(x + 1)[/tex]

Solution:
[tex]-5x^{2} + x + 6 =[/tex]
[tex]-5x^{2} -5x + 6x + 6 =[/tex]
[tex]-5x \times x - 5x \times 1 + 6 \times x + 6 \times 1 =[/tex]
[tex]-5x(x + 1) + 6(x + 1) = (x + 1)(-5x + 6)[/tex]

Solution:
[tex]x^{4} + 2x^{2} - 9x^{2} - 18 =[/tex]
[tex]x^{2} \times x^{2} + x^{2} \times 2 - 9 \times x^{2}- 9 \times 2 =[/tex]
[tex]x^{2}(x^{2} + 2) - 9(x^{2} + 2) =[/tex]
[tex](x^{2} + 2)(x^{2} - 9) =[/tex]
[tex](x^{2} + 2)(x - 3)(x + 3)[/tex]

Solution:
[tex]x^{4} + 3x^{2} - 4x^{2} - 12 =[/tex]
[tex]x^{2} \times x^{2} + x^{2} \times 3 - 4 \times x^{2} - 4 \times 3 =[/tex]
[tex]x^{2}(x^{2} + 3) - 4(x^{2} + 3) =[/tex]
[tex](x^{2} + 3)(x^{2} - 4) =[/tex]
[tex](x^{2} + 3)(x - 2)(x + 2)[/tex]

Solution:
[tex](3x + 2)^{2} + 9x + 6 =[/tex]
[tex](3x + 2)^{2} + 3(3x + 2) =[/tex]
[tex](3x + 2)[(3x + 2) + 3] =[/tex]
[tex](3x + 2)(3x + 2 + 3) =[/tex]
[tex](3x + 2)(3x + 5)[/tex]

Solution:
[tex](2x - 5)^{2} + 4x - 10 =[/tex]
[tex](2x - 5)^{2} + 2(2x - 5) =[/tex]
[tex](2x - 5)[(2x - 5) + 2] =[/tex]
[tex](2x - 5)(2x - 5 + 2) =[/tex]
(2x - 5)(2x - 3)

Solution:
[tex](5x - 7)^{2} - 15x + 21 =[/tex]
[tex](5x - 7)^{2} - 3(5x - 7)=[/tex]
[tex](5x - 7)[(5x - 7)- 3]=[/tex]
[tex](5x - 7)(5x - 7 - 3)=[/tex]
[tex](5x - 7)(5x - 10) = (5x - 7)5(x - 2) = 5(5x - 7)(x - 2)[/tex]

Solution:
[tex](4x + 1)^{2} - 8x - 2 =[/tex]
[tex](4x + 1)^{2} - 2(4x + 1) =[/tex]
[tex](4x + 1)[(4x + 1) - 2] =[/tex]
[tex](4x + 1)(4x + 1 - 2) =[/tex]
[tex](4x + 1)(4x - 1)[/tex]

Solution:
[tex]4x - 6 - (2x - 3)^{2} =[/tex]
[tex]2(2x - 3) - (2x - 3)^{2} =[/tex]
[tex](2x - 3)[2 - (2x - 3)] =[/tex]
[tex](2x - 3)(2 - 2x + 3) =[/tex]
(2x - 3)(5 - 2x)

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} + 3x + 2 =[/tex]
[tex]x^{2} + x + 2x + 2 =[/tex]
[tex]x \times x + x \times 1 + 2 \times x + 2 \times 1 =[/tex]
[tex]x(x + 1) + 2(x + 1) =[/tex]
[tex](x + 1)(x + 2)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} + 6x + 8 =[/tex]
[tex]x^{2} + 2x + 4x + 8 =[/tex]
[tex]x \times x + x \times 2 + 4 \times x + 4 \times 1=[/tex]
[tex]x(x + 2) + 4(x + 2) =[/tex]
[tex](x + 2)(x + 4)[/tex]

Solution:
[tex]9x - 12 + (3x - 4)^{2} =[/tex]
[tex]3(3x - 4) + (3x - 4)^{2} =[/tex]
[tex](3x - 4)[3 + (3x - 4)]=[/tex]
[tex](3x - 4)(3 + 3x - 4)=[/tex]
(3x - 4)(3x - 1)

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} + 6x + 8 =[/tex]
[tex]x^{2} + 4x + 2x + 8 =[/tex]
[tex]x \times x + x \times 4 + 2 \times x + 2 \times 4 =[/tex]
[tex]x(x + 4) + 2(x + 4) = (x + 4)(x + 2)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} + 9x + 20 =[/tex]
[tex]x^{2} + 5x + 4x + 20 =[/tex]
[tex]x \times x + x \times 5 + 4 \times x + 4 \times 5=[/tex]
[tex]x(x + 5) + 4(x + 5) = (x + 5)(x + 4)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} - 5x + 6 =[/tex]
[tex]x^{2} - 3x - 2x + 6 =[/tex]
[tex]x \times x + x \times (-3) - 2 \times x - 2 \times (-3) =[/tex]
[tex]x(x - 3) - 2(x - 3) = (x - 2)(x - 3)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} - 8x + 15 =[/tex]
[tex]x^{2} - 3x - 5x + 15 =[/tex]
[tex]x \times x + x \times (-3) - 5 \times x -5 \times (-3) =[/tex]
[tex]x(x - 3) - 5(x - 3) = (x - 3)(x - 5)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times e = a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} - 7x + 12 =[/tex]
[tex]x^{2} -4x - 3x + 12 =[/tex]
[tex]x \times x + x \times (-4) - 3 \times x - 3 \times (-4) =[/tex]
[tex]x(x - 4) - 3(x - 4) =[/tex]
(x - 4)(x - 3)

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times[/tex] e = [tex]a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} + x - 20 =[/tex]
[tex]x^{2} - 4x + 5x - 20 =[/tex]
[tex]x \times x + x \times (-4) + 5 \times x + 5 \times (-4) =[/tex]
[tex]x(x - 4) + 5(x - 4) = (x - 4)(x + 5)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times[/tex] e = [tex]a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} - 4x - 12 =[/tex]
[tex]x^{2} - 6x + 2x - 12 =[/tex]
[tex]x \times x + x \times (-6) + 2 \times x + 2 \times (-12) =[/tex]
[tex]x(x - 6) + 2(x - 6) = (x - 6)(x + 2)[/tex]

Solution:
We can write: [tex]ax^{2} + bx + c = ax^{2} + dx + ex + c[/tex],
[tex]d \times[/tex] e = [tex]a \times c[/tex] and [tex]d + e = b[/tex]

[tex]x^{2} - x - 6 =[/tex]
[tex]x^{2} - 3x + 2x - 6 =[/tex]
[tex]x \times x + x \times (-3) + 2 \times x + 2 \times (-3) =[/tex]
[tex]x(x - 3) + 2(x - 3) = (x - 3)(x + 2)[/tex]

Solution:
[tex]2x^2-4xy+2y^2-8=2(x^2-2xy+y^2-4)=2\bigl((x-y)^2-2^2\bigr)[/tex]
[tex]=2(x-y-2)(x-y+2)[/tex]

Solution:
Group the terms:
[tex](ax-ay)-(x^2-2xy+y^2)=a(x-y)-(x-y)^2[/tex]
[tex]=(x-y)\bigl(a-(x-y)\bigr)=(x-y)(a-x+y)[/tex]

Solution:
[tex]64-x^6=8^2-(x^3)^2=(8-x^3)(8+x^3)[/tex]
Now we use the formulas for a difference and a sum of cubes:
[tex]8-x^3=(2-x)(4+2x+x^2)[/tex]
[tex]8+x^3=(2+x)(4-2x+x^2)[/tex]
[tex]64-x^6=(2-x)(2+x)(4+2x+x^2)(4-2x+x^2)[/tex]

Solution:
Group the terms:
[tex](x^4+x^3y)-(xy^3+y^4)=x^3(x+y)-y^3(x+y)=(x+y)(x^3-y^3)[/tex]
[tex]=(x+y)(x-y)(x^2+xy+y^2)[/tex]

Solution:
Write [tex]-3a=-a-2a[/tex] and group:
[tex]a^3-a-2a+2=a(a^2-1)-2(a-1)=a(a-1)(a+1)-2(a-1)[/tex]
[tex]=(a-1)(a^2+a-2)=(a-1)(a-1)(a+2)=(a-1)^2(a+2)[/tex]

Solution:
[tex]x^5+x^4-4x-4=x^4(x+1)-4(x+1)=(x+1)(x^4-4)=(x+1)(x^2-2)(x^2+2)[/tex]
So it is divisible by [tex]x^2-2[/tex]. It is not divisible by [tex]x-1[/tex] or [tex]x+2[/tex]: for [tex]x=1[/tex] its value is [tex]-6[/tex], and for [tex]x=-2[/tex] it is [tex]-12[/tex] (not [tex]0[/tex]).

Solution:
Write [tex]-5x=-2x-3x[/tex] and group:
[tex](x^2-2x)+(xy-2y)-(3x-6)=x(x-2)+y(x-2)-3(x-2)[/tex]
[tex]=(x-2)(x+y-3)[/tex]

Solution:
[tex](7n+3)^2-(n-3)^2=(7n+3-n+3)(7n+3+n-3)=(6n+6)\times 8n=48n(n+1)[/tex]
Of the two consecutive numbers [tex]n[/tex] and [tex]n+1[/tex], one is even, so [tex]n(n+1)[/tex] is divisible by [tex]2[/tex] and [tex]48n(n+1)[/tex] is divisible by [tex]96[/tex]. For [tex]n=1[/tex] the value is [tex]96[/tex], which is not divisible by [tex]7[/tex], [tex]26[/tex] or [tex]33[/tex].

Solution:
Replace the number [tex]5[/tex] by [tex]a-2b[/tex]:
[tex]5a-2ab+4b^2=(a-2b)a-2ab+4b^2=a^2-4ab+4b^2=(a-2b)^2=5^2=25[/tex]

Solution:
[tex]\dfrac{2025^2+2025}{2026^2-1}=\dfrac{2025(2025+1)}{(2026-1)(2026+1)}=\dfrac{2025\times 2026}{2025\times 2027}=\dfrac{2026}{2027}[/tex]

Solution:
[tex]150^4-149^4=(150^2-149^2)(150^2+149^2)[/tex]
Therefore [tex]\dfrac{150^4-149^4}{150^2+149^2}=150^2-149^2=(150-149)(150+149)=299[/tex]

Solution:
Since [tex]77+55=132[/tex], we use [tex]a^3+b^3=(a+b)(a^2-ab+b^2)[/tex]:
[tex]\dfrac{77^3+55^3}{132}=\dfrac{(77+55)(77^2-77\times 55+55^2)}{132}=77^2-77\times 55+55^2[/tex]
[tex]77^2-77\times 55+55^2-77\times 55=77^2-2\times 77\times 55+55^2=(77-55)^2=22^2=484[/tex]

Solution:
This is the cube of a difference [tex](a-b)^3=a^3-3a^2b+3ab^2-b^3[/tex] with [tex]a=0.1[/tex], [tex]b=5.1[/tex]:
[tex](0.1-5.1)^3=(-5)^3=-125[/tex]

Solution:
[tex]1-\dfrac{1}{k^2}=\dfrac{k^2-1}{k^2}=\dfrac{(k-1)(k+1)}{k\times k}[/tex]
So the product equals
[tex]\dfrac{1\times 3}{2\times 2}\times \dfrac{2\times 4}{3\times 3}\times \dfrac{3\times 5}{4\times 4}\times \ldots\times \dfrac{9\times 11}{10\times 10}[/tex]
Almost all factors cancel: only [tex]\frac{1}{2}[/tex] (from the first fraction) and [tex]\frac{11}{10}[/tex] (from the last one) remain:
[tex]\dfrac{1}{2}\times \dfrac{11}{10}=\dfrac{11}{20}[/tex]

Solution:
Let the number be [tex]n[/tex]; the preceding number is [tex]n-1[/tex]:
[tex]n^2-(n-1)^2=\bigl(n-(n-1)\bigr)\bigl(n+(n-1)\bigr)=2n-1[/tex]
[tex]2n-1=625 \Rightarrow 2n=626 \Rightarrow n=313[/tex]
Indeed, its last digit is [tex]3[/tex].

Solution:
[tex]a^2-b^2+ac-bc=(a-b)(a+b)+c(a-b)=(a-b)(a+b+c)=0[/tex]
Since [tex]a+b+c\gt 0[/tex] (it is the perimeter), we get [tex]a-b=0[/tex], i.e. [tex]a=b[/tex]: the triangle is isosceles. It does not have to be equilateral or right-angled, for example [tex]a=b=2[/tex], [tex]c=3[/tex].