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 April 6th, 2014, 01:47 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 find x^3 +4x =8 find x^7+ 64x^2
 April 6th, 2014, 03:28 PM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra $\displaystyle \begin{array}{rcl}x^3+4x &=& 8 \\\\ x^3 &=& 8-4x \\\\ x^6 &=& 64-64x+16x^2 \\\\ x^6+64x &=& 16x^2+64\ =\ 16(x^2+4) \\\\ x^7+64x^2 &=& 16(x^3+4x)\ =\ 16\cdot8\ =\ 128\end{array}$ Thanks from soroban, MarkFL and Denis Last edited by Olinguito; April 6th, 2014 at 03:34 PM.
 April 6th, 2014, 06:49 PM #3 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 We need to know that we need to multiply both sides of the equation by $\displaystyle x^4$ to arrive at $\displaystyle x^7\,+\,64x^2$ for the expression $\displaystyle x^7$ so $\displaystyle x^7\,+\,4x^5\,=\,8x^4\text{.}$ We also need to know that we need to multiply both sides of the equation by $\displaystyle 16x$ to arrive at $\displaystyle x^7\,+\,64x^2$ for the expression $\displaystyle 64x^2$ so $\displaystyle 16x^4\,+\,64x^2\,=\,128x\text{.}$ Since $\displaystyle x^7\,=\,8x^4\,-\,4x^5$ and $\displaystyle 64x^2\,=\,128x\,-\,16x^4$ we get $\displaystyle x^7\,+\,64x^2\,=\,-4x^5\,-\,8x^4\,+\,128x\,=\,-4x^5\,-\,16x^3\,+\,16x^3\,-\,8x^4\,-\,32x^2\,+\,32x^2\,+\,128x\,=\,-4x^2(x^3\,+\,4x)\,-\,8x(x^3\,+\,4x)\,+\,16x^3\,+\,32x^2\,+\,128x\,=\,-32x^2\,-\,64x\,+\,16x^3\,+\,32x^2\,+\,128x\,=\,16x^3\,+\,6 4x\,=\,16(x^3\,+\,4x)\,=\,16\,\times\,8\,=\,\boxed {128}\text{.}$ Thanks from Olinguito

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