Asked by Bongiwe Gamede
a) 5(1/5x - 1)(5x - 1/5)
b)(2a - 3)^2 - (3a + 4)(3a - 4) +2(3a + 1)^2
c)(x/3 - 2)(x/2 - 3)
d) -1/5x(25x - 10y) + 1/3(12x-3y)
e) 3(x-2y)^3
b)(2a - 3)^2 - (3a + 4)(3a - 4) +2(3a + 1)^2
c)(x/3 - 2)(x/2 - 3)
d) -1/5x(25x - 10y) + 1/3(12x-3y)
e) 3(x-2y)^3
Answers
Answered by
Damon
well, we do not need the distributive property if you know FOIL
I do not dare to do a) because I do not know if you mean
(1/5)x or 1/(5x) when you write 1/5x
b)
(2a-3)(2a-3) = 4 a^2 - 12 a + 9
-(3a+4)(3a-4)= -9 a^2 + 0 a +16
2(3a+1)(3a+1)= 18a^2 + 12 a + 2
add
13 a^2 + 27
etc
I do not dare to do a) because I do not know if you mean
(1/5)x or 1/(5x) when you write 1/5x
b)
(2a-3)(2a-3) = 4 a^2 - 12 a + 9
-(3a+4)(3a-4)= -9 a^2 + 0 a +16
2(3a+1)(3a+1)= 18a^2 + 12 a + 2
add
13 a^2 + 27
etc
Answered by
Damon
Now the only trinomial is at e)
(x-2y)^2 = x^2 - 4 x y + 4 y^2
now we multiply that by (3x-6y)
use distributive property
(a+b)(c+d+e)
= a(c+d+e) + b(c+d+e)
so we have
+(3x)(x^2 - 4 x y + 4 y^2)
-(6y)(x^2 - 4 x y + 4 y^2)
=
+3 x^3 - 12 x^2y + 12 xy^2
-0 x^3 - 6 x^2 y + 24 xy^2 -24y^3
---------------------------------
=3x^2 -18 x^2y + 36 xy^2 -24 y^3
get it?
(x-2y)^2 = x^2 - 4 x y + 4 y^2
now we multiply that by (3x-6y)
use distributive property
(a+b)(c+d+e)
= a(c+d+e) + b(c+d+e)
so we have
+(3x)(x^2 - 4 x y + 4 y^2)
-(6y)(x^2 - 4 x y + 4 y^2)
=
+3 x^3 - 12 x^2y + 12 xy^2
-0 x^3 - 6 x^2 y + 24 xy^2 -24y^3
---------------------------------
=3x^2 -18 x^2y + 36 xy^2 -24 y^3
get it?
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