anonymous
  • anonymous
What is the factored form of the equation? Factor completely. 24q7 – 42q4r + 36q3r2 – 63r3
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
@Callisto
anonymous
  • anonymous
@Lilith
anonymous
  • anonymous
the two middle terms are q^4r and q^3r^2. notice q^3*q^4=q^7 (the first term) and r*r^2=r^3 (the last term)

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anonymous
  • anonymous
so we have (a*q^3-b*r)(c*q^4+d*r^2)
anonymous
  • anonymous
\[(a*q^3-b*r)(c*q^4+d*r^2)=acq^7-bcq^4r+adq^3r^2-bdr^3\]
anonymous
  • anonymous
ac=24; bc=42; ad=36; bd=63
anonymous
  • anonymous
for ac and ad (24 & 36) the common factor is 12. this means a must equal 12. so c must equal 2 (12*2=a*c=24) and d must equal 3 (12*3=a*d=36)
anonymous
  • anonymous
bc and bd have a common factor of 21. this must be b. we can check this. 21*2=b*c=42; 21*3=b*d=63 so a=12; b=21; c=2; d=3;
anonymous
  • anonymous
we can finish the factoring now. \[(aq^3−br)(cq^4+dr^2)=acq^7−bcq^4r+adq^3r^2−bdr^3\] \[(12q^3−21r)(2q^4+3r^2)=24q^7−36q^4r+42q^3r^2−63r^3\]
anonymous
  • anonymous
for the first factor \[(12q^3-21r)\] a "3" can be factored out. \[(12q^3-21r)=3(4q^3-21r)\] so our final factorization is \[3(4q^3-21r)(2q^4+3r^2)\]
anonymous
  • anonymous
that should be\[3(4q^3−7r)(2q^4+3r^2)\]

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