Math Exercises & Math Problems: Application of Formulas for ... - Free Printable

Let’s solve each expression one by one. We’ll use algebra rules like:

- $(a + b)^2 = a^2 + 2ab + b^2$
- $(a - b)^2 = a^2 - 2ab + b^2$
- $(a + b)(a - b) = a^2 - b^2$ (difference of squares)
- For cubes: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
- Also, remember to distribute negatives carefully.

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a) $(a + 2b)^2$
= $a^2 + 2(a)(2b) + (2b)^2$
= $a^2 + 4ab + 4b^2$

b) $(2x - 3y)^2$
= $(2x)^2 - 2(2x)(3y) + (3y)^2$
= $4x^2 - 12xy + 9y^2$

c) $(10a - 9b)^2$
= $(10a)^2 - 2(10a)(9b) + (9b)^2$
= $100a^2 - 180ab + 81b^2$

d) $(5m + 2n)(5m - 2n)$ → difference of squares
= $(5m)^2 - (2n)^2$
= $25m^2 - 4n^2$

e) $(1 - a)(1 + a)$ → difference of squares
= $1^2 - a^2$
= $1 - a^2$

f) $(-a + 3b)^2$ → same as $(3b - a)^2$
= $(3b)^2 - 2(3b)(a) + a^2$
= $9b^2 - 6ab + a^2$
(or written as $a^2 - 6ab + 9b^2$ — both are correct)

g) $(-x - 4)^2$ → factor out negative? Or just expand:
= $[(-1)(x + 4)]^2 = (-1)^2(x + 4)^2 = (x + 4)^2$
= $x^2 + 8x + 16$
Alternatively:
$(-x)^2 + 2(-x)(-4) + (-4)^2 = x^2 + 8x + 16$

h) $(-3x - 5y)^2$ → same as $[-(3x + 5y)]^2 = (3x + 5y)^2$
= $(3x)^2 + 2(3x)(5y) + (5y)^2$
= $9x^2 + 30xy + 25y^2$

i) $(7u - 11y)^2$
= $(7u)^2 - 2(7u)(11y) + (11y)^2$
= $49u^2 - 154uy + 121y^2$

j) $(8c - 13d)(8c + 13d)$ → difference of squares
= $(8c)^2 - (13d)^2$
= $64c^2 - 169d^2$

k) $(a^4 - b^4)^2$
= $(a^4)^2 - 2(a^4)(b^4) + (b^4)^2$
= $a^8 - 2a^4b^4 + b^8$

l) $(6xy^2 + 7x^2y)^2$
Let’s treat it as $(A + B)^2$ where A = $6xy^2$, B = $7x^2y$
= $A^2 + 2AB + B^2$
A² = $(6xy^2)^2 = 36x^2y^4$
B² = $(7x^2y)^2 = 49x^4y^2$
2AB = 2 * 6xy² * 7x²y = 84 x³ y³
So total: $36x^2y^4 + 84x^3y^3 + 49x^4y^2$

m) $-(-6u - 3v)(6u + 3v)$
First, note that $(-6u - 3v) = -(6u + 3v)$
So: $- [ -(6u + 3v) ] (6u + 3v) = - [ - (6u + 3v)^2 ] = (6u + 3v)^2$
Now expand:
= $(6u)^2 + 2(6u)(3v) + (3v)^2$
= $36u^2 + 36uv + 9v^2$

n) $(-uv + 7a)^2$ → same as $(7a - uv)^2$
= $(7a)^2 - 2(7a)(uv) + (uv)^2$
= $49a^2 - 14auv + u^2v^2$

o) $(-s + 10t)(-s - 10t)$ → this is $[(-s) + 10t][(-s) - 10t]$ → difference of squares
= $(-s)^2 - (10t)^2$
= $s^2 - 100t^2$

p) $(-5 - 3t)(-5 + 3t)$ → again, difference of squares: $[(-5) - 3t][(-5) + 3t]$
= $(-5)^2 - (3t)^2$
= $25 - 9t^2$

q) $(12u - 14y)^2$
= $(12u)^2 - 2(12u)(14y) + (14y)^2$
= $144u^2 - 336uy + 196y^2$

r) $(3x + 2y)^3$
Use cube formula: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$
a = 3x, b = 2y
= $(3x)^3 + 3(3x)^2(2y) + 3(3x)(2y)^2 + (2y)^3$
= $27x^3 + 3*9x^2*2y + 3*3x*4y^2 + 8y^3$
= $27x^3 + 54x^2y + 36xy^2 + 8y^3$

s) $(5a - 4b)^3$
Formula: $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
a = 5a, b = 4b
= $(5a)^3 - 3(5a)^2(4b) + 3(5a)(4b)^2 - (4b)^3$
= $125a^3 - 3*25a^2*4b + 3*5a*16b^2 - 64b^3$
= $125a^3 - 300a^2b + 240ab^2 - 64b^3$

t) $27a^3 - 8b^3$ → this is difference of cubes!
Recall: $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$
Here, $27a^3 = (3a)^3$, $8b^3 = (2b)^3$
So: $(3a)^3 - (2b)^3 = (3a - 2b)((3a)^2 + (3a)(2b) + (2b)^2)$
= $(3a - 2b)(9a^2 + 6ab + 4b^2)$

u) $64x^6 + 125y^9$ → sum of cubes!
$64x^6 = (4x^2)^3$, since $(4x^2)^3 = 64x^6$
$125y^9 = (5y^3)^3$, since $(5y^3)^3 = 125y^9$
Sum of cubes: $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
So: $(4x^2 + 5y^3)((4x^2)^2 - (4x^2)(5y^3) + (5y^3)^2)$
= $(4x^2 + 5y^3)(16x^4 - 20x^2y^3 + 25y^6)$

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Final Answer:

a) $a^2 + 4ab + 4b^2$
b) $4x^2 - 12xy + 9y^2$
c) $100a^2 - 180ab + 81b^2$
d) $25m^2 - 4n^2$
e) $1 - a^2$
f) $a^2 - 6ab + 9b^2$
g) $x^2 + 8x + 16$
h) $9x^2 + 30xy + 25y^2$
i) $49u^2 - 154uy + 121y^2$
j) $64c^2 - 169d^2$
k) $a^8 - 2a^4b^4 + b^8$
l) $36x^2y^4 + 84x^3y^3 + 49x^4y^2$
m) $36u^2 + 36uv + 9v^2$
n) $49a^2 - 14auv + u^2v^2$
o) $s^2 - 100t^2$
p) $25 - 9t^2$
q) $144u^2 - 336uy + 196y^2$
r) $27x^3 + 54x^2y + 36xy^2 + 8y^3$
s) $125a^3 - 300a^2b + 240ab^2 - 64b^3$
t) $(3a - 2b)(9a^2 + 6ab + 4b^2)$
u) $(4x^2 + 5y^3)(16x^4 - 20x^2y^3 + 25y^6)$