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Using the binomial theorem how would you simplify (sqrt(x) +5)^3?..........sqrt= squareroot

 Apr 6, 2016
 #1
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You can expand as follows:

Expand the following:
(sqrt(x)+5)^3

(5+sqrt(x))^3  =  sum_(k=0)^3 binomial(3, k) (sqrt(x))^(3-k) 5^k  =  binomial(3, 0) (sqrt(x))^3 5^0+binomial(3, 1) (sqrt(x))^2 5^1+binomial(3, 2) (sqrt(x))^1 5^2+binomial(3, 3) (sqrt(x))^0 5^3:
binomial(3, 0) x^(3/2)+5 binomial(3, 1) x+25 binomial(3, 2) sqrt(x)+125 binomial(3, 3)

binomial(3, 0) = 1, binomial(3, 1) = 3, binomial(3, 2) = 3 and binomial(3, 3) = 1:
5^3+5^2×3 sqrt(x)+5×3 (sqrt(x))^2+(sqrt(x))^3

5^2 = 25:
5^3+25×3 sqrt(x)+5×3 (sqrt(x))^2+(sqrt(x))^3

Cancel exponents. (sqrt(x))^2 = x:
5^3+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5^3 = 5×5^2:
5×5^2+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5^2 = 25:
5×25+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

5×25  =  125:
125+25×3 sqrt(x)+5×3 x+(sqrt(x))^3

25×3  =  75:
125+75 sqrt(x)+5×3 x+(sqrt(x))^3

5×3  =  15:
125+75 sqrt(x)+15 x+(sqrt(x))^3

Multiply exponents. (sqrt(x))^3 = x^(3/2):
Answer: |  125+75 sqrt(x)+15 x+x^(3/2)

 Apr 6, 2016
 #2
avatar+26400 
+5

Using the binomial theorem how would you simplify (sqrt(x) +5)^3?..........sqrt= squareroot

 

\(\begin{array}{lcl} (\sqrt{x} +5)^3 &=& \binom30 \cdot (\sqrt{x})^3 \cdot 5^0\\ && + \binom31 \cdot (\sqrt{x})^2 \cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1 \cdot 5^2 \\ && + \binom33 \cdot (\sqrt{x})^0 \cdot 5^3 \\\\ &=& \binom30 \cdot (\sqrt{x})^3 \cdot 1 \\ && + \binom31 \cdot (\sqrt{x})^2 \cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1 \cdot 5^2 \\ && + \binom33 \cdot 1 \cdot 5^3 \\\\ &=& \binom30 \cdot (\sqrt{x})^3 \\ && + \binom31 \cdot (\sqrt{x})^2\cdot 5^1 \\ && + \binom32 \cdot (\sqrt{x})^1\cdot 5^2 \\ && + \binom33 \cdot 5^3 \\\\ &=& \binom30 \cdot x^{ \frac{3}{2} } \\ && + \binom31 \cdot x^{ \frac{2}{2} } \cdot 5^1 \\ && + \binom32 \cdot x^{ \frac{1}{2} } \cdot 5^2 \\ && + \binom33 \cdot 5^3 \\\\ &=& \binom30 \cdot x^{ \frac{3}{2} } + \binom31 \cdot x \cdot 5^1 + \binom32 \cdot x^{ \frac{1}{2} } \cdot 5^2 + \binom33 \cdot 5^3 \end{array} \)

 

\(\begin{array}{lcl} \binom30 = \binom33 = 1 \\ \binom31 = \binom32 = 3 \end{array}\)

 

\(\begin{array}{lcl} (\sqrt{x} +5)^3 &=& 1 \cdot x^{ \frac{3}{2} } + 3 \cdot x \cdot 5^1 + 3 \cdot x^{ \frac{1}{2} } \cdot 5^2 + 1 \cdot 5^3 \\\\ &=& x^{ \frac{3}{2} } + 15 \cdot x + 75 \cdot x^{ \frac{1}{2} } + 125 \\\\ \mathbf{(\sqrt{x} +5)^3} & \mathbf{=}& \mathbf{ x^{1.5} + 15 \cdot x + 75 \cdot x^{0.5} + 125 } \end{array}\)

 

laugh

 Apr 6, 2016

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