12 multiple-choice questions, progressively harder.
Use the difference of squares to compute 43×3743 \times 3743×37.
Solution
Correct answer: C
Write each factor as a distance from 404040: 43=40+343 = 40 + 343=40+3 and 37=40−337 = 40 - 337=40−3.
43×37=(40+3)(40−3)=402−32=1600−9=159143 \times 37 = (40 + 3)(40 - 3) = 40^2 - 3^2 = 1600 - 9 = 159143×37=(40+3)(40−3)=402−32=1600−9=1591
The product is a square minus 999.
Factor x4−256x^4 - 256x4−256 completely.
Correct answer: B
Read x4−256x^4 - 256x4−256 as (x2)2−162(x^2)^2 - 16^2(x2)2−162. The factor x2−16x^2 - 16x2−16 factors again.
x4−256=(x2+16)(x2−16)=(x2+16)(x+4)(x−4)x^4 - 256 = (x^2 + 16)(x^2 - 16) = (x^2 + 16)(x + 4)(x - 4)x4−256=(x2+16)(x2−16)=(x2+16)(x+4)(x−4)
The sum of squares x2+16x^2 + 16x2+16 stops there.
In the factorization x4−16=(x2+4)(x+2)(x−2)x^4 - 16 = (x^2 + 4)(x + 2)(x - 2)x4−16=(x2+4)(x+2)(x−2), which factor is a sum of squares?
A sum of squares is a plus sign between two perfect squares. Among the factors, only x2+4x^2 + 4x2+4 has that form.
x2+4=(x)2+22x^2 + 4 = (x)^2 + 2^2x2+4=(x)2+22
A sum of squares does not factor over the reals, which is why the factoring stops there.
Use the difference of squares to compute 61×5961 \times 5961×59.
Correct answer: D
Write each factor as a distance from 606060: 61=60+161 = 60 + 161=60+1 and 59=60−159 = 60 - 159=60−1.
61×59=(60+1)(60−1)=602−12=3600−1=359961 \times 59 = (60 + 1)(60 - 1) = 60^2 - 1^2 = 3600 - 1 = 359961×59=(60+1)(60−1)=602−12=3600−1=3599
The product is a square minus 111.
Factor 50−2x250 - 2x^250−2x2 completely.
Correct answer: A
Pull out the common factor 222 first, then factor the difference of squares.
50−2x2=2(25−x2)=2(5+x)(5−x)50 - 2x^2 = 2(25 - x^2) = 2(5 + x)(5 - x)50−2x2=2(25−x2)=2(5+x)(5−x)
Stopping at 2(25−x2)2(25 - x^2)2(25−x2) is incomplete.
Factor 9x2−64y29x^2 - 64y^29x2−64y2.
The roots are 9x2=3x\sqrt{9x^2} = 3x9x2=3x and 64y2=8y\sqrt{64y^2} = 8y64y2=8y.
9x2−64y2=(3x)2−(8y)2=(3x+8y)(3x−8y)9x^2 - 64y^2 = (3x)^2 - (8y)^2 = (3x + 8y)(3x - 8y)9x2−64y2=(3x)2−(8y)2=(3x+8y)(3x−8y)
Each root keeps its coefficient with its variable.
Use the difference of squares to compute 25×1525 \times 1525×15.
Write each factor as a distance from 202020: 25=20+525 = 20 + 525=20+5 and 15=20−515 = 20 - 515=20−5.
25×15=(20+5)(20−5)=202−52=400−25=37525 \times 15 = (20 + 5)(20 - 5) = 20^2 - 5^2 = 400 - 25 = 37525×15=(20+5)(20−5)=202−52=400−25=375
The product is a square minus 252525.
Which expression equals (6x+5)(6x−5)(6x + 5)(6x - 5)(6x+5)(6x−5)?
The product (6x+5)(6x−5)(6x + 5)(6x - 5)(6x+5)(6x−5) is a difference of squares, so the middle terms cancel.
(6x+5)(6x−5)=36x2−25(6x + 5)(6x - 5) = 36x^2 - 25(6x+5)(6x−5)=36x2−25
There is no middle term, so it is not 36x2−60x+2536x^2 - 60x + 2536x2−60x+25.
Factor 3x4−33x^4 - 33x4−3 completely.
Pull out the common factor 333 first, then factor completely. The factor x2−1x^2 - 1x2−1 factors again.
3x4−3=3(x4−1)=3(x2+1)(x+1)(x−1)3x^4 - 3 = 3(x^4 - 1) = 3(x^2 + 1)(x + 1)(x - 1)3x4−3=3(x4−1)=3(x2+1)(x+1)(x−1)
The sum of squares x2+1x^2 + 1x2+1 does not factor further.
Solve x2−121=0x^2 - 121 = 0x2−121=0.
Factor and use the zero-product property.
x2−121=(x+11)(x−11)=0 ⟹ x=11 or x=−11x^2 - 121 = (x + 11)(x - 11) = 0 \implies x = 11 \text{ or } x = -11x2−121=(x+11)(x−11)=0⟹x=11 or x=−11
Both roots are solutions.
Use the difference of squares to compute 55×4555 \times 4555×45.
Write each factor as a distance from 505050: 55=50+555 = 50 + 555=50+5 and 45=50−545 = 50 - 545=50−5.
55×45=(50+5)(50−5)=502−52=2500−25=247555 \times 45 = (50 + 5)(50 - 5) = 50^2 - 5^2 = 2500 - 25 = 247555×45=(50+5)(50−5)=502−52=2500−25=2475
Factor 3x2−753x^2 - 753x2−75 completely.
Pull out the common factor 333 first, then factor the difference of squares.
3x2−75=3(x2−25)=3(x+5)(x−5)3x^2 - 75 = 3(x^2 - 25) = 3(x + 5)(x - 5)3x2−75=3(x2−25)=3(x+5)(x−5)
Stopping at 3(x2−25)3(x^2 - 25)3(x2−25) is incomplete.
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