QUESTION IMAGE
Question
- simplify the expression below.
$(9m^3n^6)^{-2} cdot (-6m^2n)^4$
a. $dfrac{m^2}{16n^8}$ c. $dfrac{16m^2}{n^8}$
b. $dfrac{4m^2}{3n^8}$ d. $dfrac{16}{m^{48}n^{48}}$
- simplify the expression below.
$2x^6y^2(4x^2y^3 - 3y) + 3x^8y^5$
a. $11x^{16}y^{10} - 6x^6y^3$
b. $11x^8y^5 - 6x^6y^3$
c. $8x^{12}y^6 + 3x^8y^5 - 6x^6y^2$
d. $5x^{14}y^8$
- simplify the expression below.
$(4y + 1)^3$
a. $12y^3 + 1$
b. $64y^3 + 1$
c. $64y^3 + 48y^2 + 16y + 3$
d. $64y^3 + 48y^2 + 12y + 1$
- completely factor the expression below.
$c^4 - c^2 - 72$
- completely factor the expression below.
$8w^3 + 125$
- completely factor the expression below.
$48m^5n - 3mn^5$
Problem 1
Step1: Apply negative exponent rule
$(9m^3n^6)^{-2} = \frac{1}{(9m^3n^6)^2} = \frac{1}{81m^6n^{12}}$
Step2: Expand the 4th power term
$(-6m^2n)^4 = 1296m^8n^4$
Step3: Multiply the two expressions
$\frac{1}{81m^6n^{12}} \cdot 1296m^8n^4 = \frac{1296}{81} \cdot \frac{m^8}{m^6} \cdot \frac{n^4}{n^{12}} = 16m^2n^{-8} = \frac{16m^2}{n^8}$
Problem 2
Step1: Distribute the first term
$2x^4y^2(4x^2y^3 - 3y) = 8x^{6}y^{5} - 6x^4y^3$
Step2: Combine like terms
$8x^{6}y^{5} - 6x^4y^3 + 3x^6y^5 = 11x^6y^5 - 6x^4y^3$
Problem 3
Step1: Use binomial cube formula
$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$, where $a=4y, b=1$
Step2: Substitute and calculate
$(4y)^3 + 3(4y)^2(1) + 3(4y)(1)^2 + 1^3 = 64y^3 + 48y^2 + 12y + 1$
Problem 4
Step1: Substitute $u=c^2$
$c^4 - c^2 -72 = u^2 - u -72$
Step2: Factor quadratic in $u$
$u^2 - u -72 = (u-9)(u+8)$
Step3: Substitute back $u=c^2$
$(c^2 - 9)(c^2 + 8) = (c-3)(c+3)(c^2+8)$
Problem 5
Step1: Recognize sum of cubes
$8w^3+125=(2w)^3+5^3$
Step2: Apply sum of cubes formula
$a^3+b^3=(a+b)(a^2-ab+b^2)$, so $(2w+5)(4w^2-10w+25)$
Problem 6
Step1: Factor out GCF
$48m^5n - 3mn^5 = 3mn(16m^4 - n^4)$
Step2: Factor difference of squares
$16m^4 - n^4=(4m^2-n^2)(4m^2+n^2)$
Step3: Factor remaining difference of squares
$4m^2-n^2=(2m-n)(2m+n)$, so $3mn(2m-n)(2m+n)(4m^2+n^2)$
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- C. $\frac{16m^2}{n^8}$
- B. $11x^6y^5 - 6x^4y^3$
- D. $64y^3 + 48y^2 + 12y + 1$
- $(c-3)(c+3)(c^2+8)$
- $(2w+5)(4w^2-10w+25)$
- $3mn(2m-n)(2m+n)(4m^2+n^2)$