Week 4 Flashcards
-5m(2n - 3m) … Can variables be distributed when simplifying?
Yes: -10mn + 15m^2
(6c + 12d) - (10c - 5d) Solution to coefficient of “c” signed or unsigned? Why?
Subtraction is not Commutative thus: (6c) - (10c) = -4c
(-12) - (-4) means: (+4) OR (-4) ?
+4 double - turns + (-12) + (-4) is the same as (-4)
2n(3n+ 1)(4n+ 2) make into quadratic
Expand the brackets and times by 2n as a whole after = 2n[12n^2+ 6n+ 4n+ 2] = 2n[12n^2+ 10n+ 2] = 24n^3+ 20n^2+ 4n Also can mult by 2n first
(2c+ 3)(c−1)(c+ 2) Use foil
Foil first two then foil for remaining two = (2c^2−2c+ 3c−3)(c+ 2) = (2c^2+c−3)(c+ 2) = 2c^3+ 4c^2+c^2+ 2c−3c−6 = 2c^3+ 5c^2−c−6
13x−(5x−2)(1−x)
keep parans until fully resolved arrange parans into quadratic = 13x −(5x −5x^2 −2 + 2x) = 13x −(−5x^2 + 7x −2) = 13x + 5x^2 −7x + 2 = 5x^2 + 6x + 2
(√7y)(√7y)
Root is negated variable stacks = 7y^2
-√7y -√7y
Surds stack -2√7y
(4−2a)(4 + 2a) does foil work?
Difference of two squares Foil only gets incomplete answer, must add the index if foil doesnt get that = 4^2−(2a)^2 = 16−4a^2
−(x−9)(x+ 9) What does - do?
Minus is applied to the signing of each term. = −(x^2−9^2) = −(x^2−81) = −x^2+81
1960 Prime factorise
repeated division of lowest factor 2 | 1960 2 | 980 2 | 490 5 | 245 7 | 49 7 | 7 1 stop at prime
HCF? 2 | 1960 2 | 1512 2 | 980 2 | 756 2 | 490 2 | 378 5 | 245 3 | 189 7 | 49 3 | 63 7 | 7 3 | 21 1 7 | 7 1
product of like factors 2^3 * 7 = 56
4a^3+12a^2 factorise
4a^2(a+3)
8-2n^2 factorise
2(4-n^2) factor out 2(2-n)(2+n) DOTS
0.01-x^2 factorise
0.1^2 = 0.01 thus (0.1-x)(0.1+x)
4nm^2-256n factorise
4n(m^2-64) factor out 4n(m-8)(m+8) DOTS
11-k^2 factorise
root to surd (√11+k)(√11-k)
2n^2+14n+20 factorise
Before using crossfire first the coefficient must be factored out 2(n^2+7n+10) then crossfire 2(n+5)(n+2)
-x^2-8x-12 factorise
-(x^2-8x-12) factor out crossfire since theres subtraction at head, the answers +6 & +2 can be used with the sub preserved: -(x+6)(x+2)
4m^2+2m-6 factorise
2(2m^2+m-3) crossfire 2(2m+3)(m-1)
y^4−28y^2+ 75 factorise
To make this easier to factorise, let A=y^2 = A^2−28A+ 75 substitute = (A−3)(A−25) crossfire = (y^2−3)(y^2−25) unsubstitute = (y^2−3)(y+ 5)(y−5) DOTS
(a+ 3)^2+ 8(a+ 3) + 12
= B^2+ 8B+ 12 = (B+ 2)(B+ 6) = ((a+ 3) + 2)((a+ 3) + 6) = (a+ 5)(a+ 9)
y-1/5 + y+3/2
= 2y-2/10 + 5y+15/10 = 7y+13/10
2c/(c+5) + 4/(c-1)
Q. 2c/(c+5) + 4/(c-1) = 2c(c-1)/(c+5)(c-1) + 4(c+5)/(c+5)(c-1) = 2c(c-1)+4(c+5)/(c+5)(c-1) = 2c^2-2c+4c+20/(c-1)(c+5) = 2c^2+2c+20/(c-1)(c+5) This is complete, quad wont factorise further



















