Simplifying Polynomial Expressions: Difficult Problems with Solutions

Solution:
$-\dfrac{6}{x^{-5}}$ because $-\dfrac{6}{x^{-5}} =-6(x^{-5})^{-1}=-6x^{5}$
$\dfrac{6}{-x^{5}} = -6x^{-5}$
Exponent of the variable has to be a nonnegative integer.

Solution:
Monomial because we add up two like terms.
$-2x^{4}+5x^{4}= 3x^{4}$

Solution:
2x²y - 17x²y = -15x²y

Solution:
$-8xy^{3}-12xy^{3}=-20xy^{3}$

Solution:
$-5x^{6}+(-4x^{6})=-5x^{6}-4x^{6}=-9x^{6}$

Solution:
$14x^{4}-(-9x^{4})=14x^{4}+9x^{4}=23x^{4}$

Solution:
$3\cdot(-2x^{4})-5\cdot(-4x^{4})= -6x^{4} + 20x^{4} = 14x^{4}$

Solution:
$4x^{2}y^{3}\cdot3xy=4\cdot3x^{2+1}y^{3+1}=12x^{3}y^{4}$

Solution:
$-3x^{4}y^{7}\cdot(-4)x^{5}y^{6}=(-3)\cdot(-4)x^{4+5}y^{7+6}=12x^{9}y^{13}$

Solution:
$A=L\cdot W=3x^{2}\cdot5x^{3}=15x^{5}$

Solution:
$12x^{4}y^{7}\div4x^{2}y^{4}=(12\div4)x^{4-2}y^{7-4}=3x^{2}y^{3}$

Solution:
$ -16x^{14}y^{8}\div(-8)x^{7}y^{5}=[(-16)\div(-8)]x^{14-7}y^{8-5}=2x^{7}y^{3}$

Solution:
$42x^{10}y^{5}\div(-6)x^{7}y^{5}=$
$[(42)\div(-6)]x^{10-7}y^{5-5}=-7x^{3}y^{0}=-7x^{3}$

Solution:
$(2x^{4}y^{3})^{5}=2^{5}x^{4\cdot5}y^{3\cdot5}=32x^{20}y^{15}$

Solution:
$(-5x^{2}y^{4})^{2}=(-5)^{2}x^{2\cdot2}y^{4\cdot2}=25x^{4}y^{8}$

Solution:
$A=l^{2}=(3x^{2})^{2}=9x^{4}$

Solution:
$V=l^{3}=(3x^{4})^{3}=27x^{12}$

Solution:
$[AC]= [AB]+[BC]=(3x^{2}+1)+(5x^{2}-4)=$
$3x^{2}+1+5x^{2}-4 =8x^{2}-3$

Solution:
$[BC]=[AC]-[AB] = (7x^{2}-3) -(3x^{2}+1) =$
$ 7x^{2}-3-3x^{2}-1 =$
$4x^{2}-4$

Solution:
$[AB]+[AC]+[BC] = (2x^{2}+5)+(6-4x)+(x+4x^{2}-9)= $
$ 2x^{2}+5+6-4x+x+4x^{2}-9=$
$ 6x^{2}-3x+2$

Solution:
$P=4\cdot l=4\cdot(2x^{2}-3x+4)=8x^{2}-12x+16$

Solution:
$P=2L+2W=2\cdot(4x+7)+2\cdot(9-2x)=$
$8x+14+18-4x=4x+32$

Solution:
$P_{1}(x)+P_{2}(x)=(3x^{2}-7x+5)+(-5x^{2}-4x+2)=$
$3x^{2}-7x+5-5x^{2}-4x+2=$
$-2x^{2}-11x+7$

Solution:
$P_{1}(x)-P_{2}x=(-4x^{3}+2x^{2}+3)-(-2x^{3}+7x^{2}-6)=$
$-4x^{3} + 2x^{2} + 3+ 2x^{3} - 7x^{2} +6 =$
$-2x^{3} - 5x^{2} +9$

Solution:
$P_{1}(x)=3x^{3}-2x^{2}-4x +5$
$P_{2}(x)=-2x^{3}+3x^{2}-2x +1$
$3P_{1}(x)-2P_{2}(x)=$
$3\cdot(3x^{3}-2x^{2}-4x +5)-2\cdot(-2x^{3}+3x^{2}-2x +1)=$
$9x^{3}-6x^{2}-12x + 15+4x^{3}-6x^{2}+4x-2 =$
$13x^{3}-12x^{2}-8x+13$

Solution:
$5x\cdot(2x^{2}-3x)-3x^{2}\cdot(-6x^{3}+4x^{2}+2x -1)=$
$10x^{3}-15x^{2} + 18x^{5}-12x^{4}-6x^{3}+3x^{2} =$
$18x^{5}-12x^{4}+4x^{3}-12x^{2}$

Solution:
$A = L\cdot W = (4x+3)(3x-5)=4x\cdot(3x-5)+3\cdot(3x-5)=$
$4x\cdot3x+4x\cdot(-5)+3\cdot3x+3\cdot(-5)=$
$12x^{2}-20x+9x-15=$
$12x^{2}-11x-15$

Solution:
$V=L\cdot W\cdot H=$
$(3x+5)\cdot(x-2)\cdot x^{2}=$
$[3x\cdot(x-2)+5\cdot(x-2)]\cdot x^{2}=$
$(3x^{2}-6x+5x-10)\cdot x^{2}=$
$(3x^{2}-x-10)\cdot x^{2}=$
$3x^{2}\cdot x^{2}-x\cdot x^{2}-10\cdot x^{2}=$
$3x^{4}-x^{3}-10x^{2}$

Solution:
We use the formula of the square of a binomial:
$(a+b)^{2}=a^{2}+2\cdot a\cdot b + b^{2}$

$(3x^{2}+4)^{2}=$
$(3x^{2})^{2}+2\cdot 3x^{2}\cdot4 +4^{2}=$
$9x^{4}+24x^{2}+16$

Solution:
We use the formula of the square of a binomial:
$(a-b)^{2}=a^{2}-2\cdot a\cdot b + b^{2}$

$(2x^{2}-1)^{2}=$
$(2x^{2})^{2}-2\cdot 2x^{2}\cdot1 +1^{2}=$
$4x^{4}-4x^{2}+1$

Solution:
We use the product of the sum and the difference of the two terms:
$(a+b)\cdot(a-b)=a^{2}-b^{2}$

$(x^{2}+6)\cdot(x^{2}-6)=(x^{2})^{2}-6^{2}=x^{4}-36$

Solution:
We use the product of the sum and the difference of the two terms:
(a + b)⋅(a - b) = a² - b²=

(x³ + 1)⋅(x³ - 1) = (x³)² - 1² = x⁶ - 1

Solution:
We use the product of the sum and the difference of the two terms:
(a + b)⋅(a - b) = a² - b²

(9x² + 4)⋅(9x² - 4) = (9x²)² - 4² = 81x⁴ - 16

Solution:
(2x - 3)(4x + 1) + (3x + 5)(x - 2) =
[2x(4x + 1) - 3(4x + 1) + 3x(x - 2) + 5(x - 2)] =
(8x² + 2x - 12x - 3 + 3x² - 6x + 5x - 10) =
11x² - 11x - 13

Solution:
(5x - 2)(3x + 4) - (2x + 7)(x + 3) =
{5x(3x + 4) - 2(3x + 4) -[2x(x + 3) + 7(x + 3)]} =
[15x² + 20x - 6x - 8 - (2x² + 6x + 7x + 21)]=
(15x² + 20x - 6x - 8 - 2x² - 6x - 7x - 21) =
13x² +x - 29

Solution:
We use the formula of the square of a binomial:
(a + b)² = a² + 2⋅a⋅b + b²

(x - 3)² - (x + 4)² =
(x² - 6x + 9) - (x² + 8x + 16) =
x²- 6x + 9 - x² - 8x - 16 =
-14x - 7

Solution:
We use the formula of the square of a binomial:
(a + b)² = a² + 2⋅a⋅b + b²

(2x + 5)² - (3x - 4)² =
[(2x)² + 2⋅2x⋅5 + 5²] - [(3x)² - 2⋅3x⋅4 + 4²] =
(4x² + 20x + 25) - (9x² - 24x + 16 =
4x² + 20x + 25 - 9x² + 24x - 16 =
-5x² + 44x + 9

Solution:
We use the formula of the square of a binomial:
(a + b)² = a² + 2⋅a⋅ b + b²
(2x + 1)² + (x - 2)²=
[(2x)² + 2˙2x⋅1 + 1] + (x² - 2⋅x⋅2 + 2²) =
4x² + 4x + 1 + x² - 4x + 4 =
5x² + 5

Solution:
We use the product of the sum and the difference of the two terms:
(a + b)⋅(a - b)=a² - b²

(x - 1)(x + 1) - (x - 3)(x + 3) =
(x² - 1²) - (x² - 3²) =
(x² - 1) - (x² - 9) =
x² - 1 - x² + 9 = 8

Solution:
We use the product of the sum and the difference of the two terms:
(a + b)⋅(a - b) = a² - b²=

(2x - 3)(2x + 3) - (3x - 5)(3x + 5)=
[(2x)² - 3²] - [(3x)² - 5²] =
(4x² - 9) - (9x² - 25) =
4x² - 9 - 9x² + 25 =
-5x² + 16

Solution:
(4x - 7)(4x + 7) + (2x - 9)(2x + 9) =
[(4x)² - 7²] + [(2x)² - 9²] =
(16x² - 49) + (4x² - 81) =
16x² - 49 + 4x² - 81 =
20x² - 130