Algebra Pre-Algebra and Basic Algebra Math Forum

 April 6th, 2014, 12:47 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 find x^3 +4x =8 find x^7+ 64x^2 April 6th, 2014, 02:28 PM #2 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 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 02:34 PM. April 6th, 2014, 05:49 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 Tags find Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post kingkos Algebra 5 November 24th, 2012 11:40 PM kevinman Algebra 8 March 8th, 2012 05:53 AM

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