Simplifying Polynomial Expressions: Problems with Solutions

Solution:
[tex]3x^{2}[/tex] because the exponent of the variable has to be a nonnegative integer.

Solution:
[tex]-4x^{3}[/tex] because [tex]2x^{2}-1[/tex]
is formed by two unlike monomials.
[tex]2x^{2} - 1[/tex] is a binomial.

Solution:
[tex]4x^{7}[/tex] because the exponent of the variable has to be a nonnegative integer.

Solution:
$-7x^{5}$ because $ \dfrac{4}{x^{3}}$ =$4x^{-3}$
Exponent of the variable has to be a nonnegative integer.

Solution:
Yes, because they have the same variables with the same exponent.

Solution:
No, because the exponents of the variables are not the same.

Solution:
No, because they have different variables: [tex]x[/tex] and [tex]y[/tex].

Solution:
Yes, because they have the same variables with the same exponents; only the coefficients [tex]-8[/tex] and [tex]7[/tex] differ.

Solution:
No: the exponents of [tex]a[/tex] and [tex]b[/tex] are different, and the first term also contains [tex]y[/tex].

Solution:
[tex]6x + 8x = (6 + 8)x = 14x[/tex]

Solution:
[tex]3x^{4} + 9x^{4} = (3 + 9)x^{4} = 12x^{4}[/tex]

Solution:
[tex]6x^{5} + 8x^{5} = 14x^{5}[/tex]

Solution:
[tex]-7x^{2} + 10x^{2} = (-7 + 10)x^{2} = 3x^{2}[/tex]

Solution:
[tex]-12x^{3} + 5x^{3} = -7x^{3}[/tex]

Solution:
[tex]-8x^{6} + 8x^{6} = (-8 + 8)x^{6} = 0x^{6} = 0[/tex]

Solution:
[tex]9x^{6} - 3x^{6}=(9 - 3)x^{6} = 6x^{6}[/tex]

Solution:
[tex]x^{4} - 8x^{4} = (1 - 8)x^{4} = -7x^{4}[/tex]

Solution:
[tex]7a^{4} - 11a^{4} = -4a^{4}[/tex]

Solution:
[tex]-4x^{9} - 6x^{9} = (-4 - 6)x^{9} = -10x^{9}[/tex]

Solution:
[tex]-8y^{2} - 9y^{2} = -17y^{2}[/tex]

Solution:
[tex]4x^{3} \times 6 = 4 \times 6x^{3} = 24x^{3}[/tex]

Solution:
[tex]-7x^{8} \times (-3) = -7 \times (-3)x^{8} = 21x^{8}[/tex]

Solution:
[tex]5x^{3} \times (-2) = -2 \times 5x^{3} = -10x^{3}[/tex]

Solution:
[tex]8x \times 3x = 8 \times 3x^{1 + 1} = 24x^{2}[/tex]

Solution:
[tex]-9x^{2} \times (-4x) = (-9) \times (-4)x^{2+1} = 36x^{3}[/tex]

Solution:
[tex]6x^{4} \times 3x^{2} = 6 \times 3x^{4 + 2} = 18x^{6}[/tex]

Solution:
[tex]-4x^{5} \times 2x^{4} = -4 \times 2x^{5 + 4} = -8x^{9}[/tex]

Solution:
[tex]6x^{5} \times 8y^{4} = 6 \times 8x^{5}y^{4} = 48x^{5}y^{4}[/tex]

Solution:
[tex]-4x^{2} \times 7y^{3} = -4 \times 7x^{2}y^{3} = -28x^{2}y^{3}[/tex]

Solution:
[tex]48x^{3} : (-6) = [48 : (-6)]x^{3} = -8x^{3}[/tex]

Solution:
[tex]12x^{6} \div 4x^{3} = 12 : 4x^{6 - 3} = 3x^{3}[/tex]

Solution:
[tex]32x^{8} \div 8x = 32 \div 8x^{8 - 1} = 4x^{7}[/tex]

Solution:
[tex]-15x^{5} \div 5x^{4} = (-15 \div 5)x^{5 - 4} = -3x^{1} = -3x[/tex]

Solution:
[tex]-24x^{8} \div 3x^{8} = (-24 \div 3)x^{8 - 8} = -8x^{0}=-8[/tex]

Solution:
[tex](-2x)^{4} = (-2)^{4}x^{1 \times 4} = 16x^{4}[/tex]

Solution:
[tex](2x^{3})^{4} = 2^{4}x^{3 \times 4} = 16x^{12}[/tex]

Solution:
[tex](-3x^{3})^{3} = (-3)^{3}x^{3 \times 3} = -27x^{9}[/tex]

Solution:
[tex]5x - 3 + 4x + 7 = 5x + 4x = 3 + 7 = 9x + 4[/tex]

Solution:
[tex]3x^{2} + 9 + 7x - 4 = 3x^{2} + 7x + 9 - 4 = 3x^{2} + 7x + 5[/tex]

Solution:
[tex]-5x^{3} + 8x - 5 - 3x^{3} - 4x + 1=[/tex]
[tex]-5x^{3} -3x^{3} + 8x - 4x - 5 + 1 =[/tex]
[tex]-8x^{3} + 4x - 4[/tex]

Solution:
[tex]6x^{3} - 4x^{2} + 5x - 1 - 3x^{2} + 4 =[/tex]
[tex]6x^{3} - 4x^{2} - 3x^{2} + 5x - 1 + 4 =[/tex]
[tex]6x^{3} - 7x^{2} + 5x + 3[/tex]

Solution:
[tex](6x^{2} - 5x + 3) - (4x^{2} + 7x + 8) =[/tex]
[tex]6x^{2} - 5x + 3 - 4x^{2} - 7x - 8 =[/tex]
[tex]2x^{2} - 12x - 5[/tex]

Solution:
[tex]4 \times (3x - 5) = 4 \times 3x + 4 \times (-5) = 12x - 20[/tex]

Solution:
[tex]-8 \times (2x^{2} - 7x + 3)=[/tex]
[tex]-8 \times 2x^{2} + (-8) \times (-7)x + (-8) \times 3=[/tex]
[tex]-16x^{2} + 56x - 24[/tex]

Solution:
[tex](-5x^{2} + 4x - 1) \times 6 =[/tex]
[tex]6 \times (-5)x^{2} + 6 \times 4x + 6 \times (-1) =[/tex]
[tex]-30x^{2} + 24x - 6[/tex]

Solution:
[tex]3x \times (2x - 5)=3x \times 2x + 3x \times (-5)=6x^{2} - 15x[/tex]

Solution:
[tex](-8x^{2} + 7x - 3) \times 9x=[/tex]
[tex]9x \times (-8)x^{2} + 9x \times 7x + 9x \times (-3) =[/tex]
[tex]-72x^{3} + 63x^{2} - 27x[/tex]

Solution:
[tex]-5x^{3} \times (-2x^{2} + 3x - 4) =[/tex]
[tex]-5x^{3} \times (-2)x^{2} - 5x^{3} \times 3x - 5x^{3} \times (-4) =[/tex]
[tex]10x^{5} - 15x^{4} + 20x^{3}[/tex]

Solution:
[tex]2x^{2} \times (-4x^{3} + 2x - 5) =[/tex]
[tex]2x^{2} \times (-4)x^{3} + 2x^{2} \times 2x + 2x^{2} \times (-5) =[/tex]
[tex]-8x^{5} + 4x^{3} - 10x[/tex]

Solution:
[tex](2x^{2})^{3}(3x - 2) =[/tex]
[tex]2^{3}(x^{2})^{3}(3x - 2) =[/tex]
[tex]8x^{6}(3x - 2) =[/tex]
[tex]8x^{6} \times 3x + 8x^{6} \times (-2)=[/tex]
[tex]24x^{7} - 16x^{6}[/tex]

Solution:
[tex](4x - 3) \times (2x + 7) =[/tex]
[tex]4x \times 2x + 4x \times 7 - 3 \times 2x - 3 \times 7 =[/tex]
[tex]8x^{2} + 28x - 6x - 21 = 8x^{2} + 22x - 21[/tex]

Solution:
$(3x+4)\cdot(2-5x) =$
$3x\cdot2 + 3x\cdot(-5x) +4\cdot2 +4\cdot(-5x)=$
$6x -15x^{2} + 8 - 20x =$
$-15x^{2}-14x +8$

Solution:
$(3x^{2}+8)\cdot(2x-9) =$
$ 3x^{2}\cdot2x + 3x^{2}\cdot(-9) +$
$8\cdot2x +8\cdot(-9)= $
$6x^{3}-27x^{2}+16x-72$

Solution:
$5x^{2}-(4x-3)(3x+2)=$
$5x^{2}-(4x\cdot3x+4x\cdot2-3\cdot3x-3\cdot2)=$
$5x^{2}-(12x^{2}+8x-9x-6)=$
$5x^{2}-(12x^{2}-x-6)=$
$5x^{2}-12x^{2}+x+6=$
$-7x^{2}+x+6$

Solution:
$2x^{2}+(5x+4)(2x-3)=$
$2x^{2}+[5x\cdot2x+5x\cdot(-3)+4\cdot2x+4\cdot(-3)]=$
$2x^{2}+(10x^{2}-15x+8x-12)=$
$2x^{2}+(10x^{2}-7x-12)=$
$2x^{2}+10x^{2}-7x-12=$
$12x^{2}-7x-12$

Solution:
$3(5x-2)(2x+3)=3(5x\cdot2x+5x\cdot3-2\cdot2x-2\cdot3) =$
$ 3(10x^{2}+15x-4x-6) =$
$ 3(10x^{2}+11x-6)=$
$30x^{2}+33x-18$

Solution:
$(4x-3)(3x-7)(-2)=[4x\cdot3x+4x\cdot(-7)-3\cdot3x-3\cdot(-7)](-2) =$
$ (12x^{2}-28x-9x+21)(-2) =$
$ (12x^{2}-37x+21)(-2)=$
$-24x^{2}+74x-42$

Solution:
We use the formula of the square of a binomial:
$(a+b)^{2}=a^{2}+2\cdot a\cdot b + b^{2}$
$(x+3)^{2}=x^{2}+2\cdot x\cdot3 +3^{2}=x^{2}+6x+9$

Solution:
We use the formula of the square of a binomial:
$(a+b)^{2}=a^{2}+2\cdot a\cdot b + b^{2}$
$(x+5)^{2}=x^{2}+2\cdot x\cdot5 +5^{2}=x^{2}+10x+25$

Solution:
We use the formula of the square of a binomial:
$(a+b)^{2}=a^{2}+2\cdot a\cdot b + b^{2}$
$(2x+7)^{2}=(2x)^{2}+2\cdot 2x\cdot7 +7^{2}=4x^{2}+28x+49$

Solution:
We use the formula of the square of a binomial:
$(a-b)^{2}=a^{2}-2\cdot a\cdot b + b^{2}$
$(x-2)^{2}=x^{2}-2\cdot x\cdot2 +2^{2}=x^{2}-4x+4$

Solution:
We use the formula of the square of a binomial:
$(a-b)^{2}=a^{2}-2\cdot a\cdot b + b^{2}$
$(x-7)^{2}=x^{2}-2\cdot x\cdot7 +7^{2}=x^{2}-14x+49$

Solution:
We use the formula of the square of a binomial:
$(a-b)^{2}=a^{2}-2\cdot a\cdot b + b^{2}$
$(5x-3)^{2}=(5x)^{2}-2\cdot 5x\cdot3+3^{2}=25x^{2}-30x+9$

Solution:
We use the product of the sum and the difference of the two terms:
$(a+b)\cdot(a-b)=a^{2}-b^{2}$
$(x+5)\cdot(x-5)=x^{2}-5^{2}=x^{2}-25$

Solution:
We use the product of the sum and the difference of the two terms:
$(a+b)\cdot(a-b)=a^{2}-b^{2}$
$(x-3)\cdot(x+3)=x^{2}-3^{2}=x^{2}-9$

Solution:
We use the product of the sum and the difference of the two terms:
$(a+b)\cdot(a-b)=a^{2}-b^{2}$
$(3x-4)\cdot(3x+4)=(3x)^{2}-4^{2}=9x^{2}-16$

Solution:
We use the product of the sum and the difference of the two terms:
$(a+b)\cdot(a-b)=a^{2}-b^{2}=$
$(5x+7)\cdot(5x-7)=(5x)^{2}-7^{2}=25x^{2}-49$

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

$(9-x)\cdot(9+x)=9^{2}-x^{2}=81-x^{2}$

Solution:
The left side of the last equality is a square, so it is never negative, while the right side [tex]-3(x-y)^2[/tex] is never positive. For example, for [tex]x=1[/tex], [tex]y=0[/tex]: [tex]9\neq -3[/tex].
The others are identities: [tex](3x-3y)^2=9(x-y)^2=9(y-x)^2[/tex] and [tex](3x+3y)^2=9(x+y)^2[/tex].

Solution:
[tex]x^4=(x^2)^2[/tex], [tex]16y^2=(4y)^2[/tex]
[tex](x^2+4y)^2=x^4+2\times x^2\times 4y+16y^2=x^4+8x^2y+16y^2[/tex]

Solution:
[tex]2y+z=3x+2x=5x[/tex]
[tex]2y-z=3x-2x=x[/tex]
[tex]2yz=3x\times 2x=6x^2[/tex]
So [tex]2x=2y+z[/tex] would mean [tex]2x=5x[/tex], which is false for [tex]x\neq 0[/tex].

Solution:
The degree of a monomial is the sum of the exponents of its variables (the coefficient [tex]27[/tex] does not count):
[tex](n+2)+5+2n=37 \Rightarrow 3n=30 \Rightarrow n=10[/tex]

Solution:
[tex]16b-24a=-8(3a-2b)=-8\times 8=-64[/tex]