function SOLNfactors(A,B,C,v,ka,a,kb,b){
var r1=new RAT(-B+Math.round(Math.sqrt(B*B-4*A*C)),2*A), r2=new RAT(-B-Math.round(Math.sqrt(B*B-4*A*C)),2*A);
if(arguments.length==4){ka=r1.bot;a=-r1.top; kb=r2.bot;b=-r2.top;};
var K=gcd(A,gcd(B,C));if(K>1){document.writeln(
showterms([[A,v,2],[B,v],[C,1]," = ",[K,1],"(",[A/K,v,2],[B/K,v],[C/K,1],")"])," and we now factor the right-hand side:
");A=A/K;B=B/K;C=C/K};
if(A<0){document.writeln("The roots are the same as those of ",showterms([[-A,v,2],[-B,v],[-C,1]]),".
");
A=-A;B=-B;C=-C};
if(C==0){document.writeln("Both terms have a factor of ",showterms([[gcd(A,B),v]]),"
",
"So the complete factorization is ",showterms([[K*gcd(A,B),v],"(",[Math.round(A/gcd(A,B)),v],[Math.round(B/gcd(A,B)),1],")"]));
return};
if(B==0 && issq(A) && (C<0) && issq(-C))
{document.writeln("Both of the terms are squares: ",showterms([[A,v,2]]),
(A!=1?showterms([" = (",[Math.round(Math.sqrt(A)),v],")2"]):""),
" and ",
showterms([[-C,1]," = ",[Math.round(Math.sqrt(-C)),1],"2"]),":
",
"So we can use the difference of two squares: ",
showterms([[1,'a'],"2 – ",[1,'b2']," = (",[1,'a'],[1,'b'],")(",[1,'a'],[-1,'b'],")"]),
"
Therefore ",showterms([[A,v,2],[B,v],[C,1]," = (",[ka,v],[a,1]," )( ",[kb,v],[b,1],")"]),
(K>1?" and "+showterms([[K*A,v,2],[K*B,v],[K*C,1]," = ",[K,1],"(",[ka,v],[a,1]," )( ",[kb,v],[b,1],")"]):"")
);
return
};
if(A==1)
{var fps=factorpairs(Math.abs(C));
document.writeln("The coefficient of ",showterms([[1,v,2]])," is ",showterms([[1,1]])," so our reference equation is",
showterms(["(",[1,v],[1,'p'],")(",[1,v],[1,'q'],") = ",[1,v,2],' + (',[1,'p'],[1,'q'],')',[1,v],[1,'pq']]),
"
We will look for two numbers ",showterms([[1,'p']])," and ",showterms([[1,'q']]),
" that
multiply to the constant term: ",showterms([[C,1]]),"",showterms([" = ",[1,'p q']])," and
",
"sum to the ", showterms([[1,v]])," coefficient: ",showterms([[B,1]]),"",showterms([" = ",[1,'p'],[1,'q']]),"" ,
(fps.length>1?"There are "+fps.length+" pairs of factors ":"There is only one pair of factors "),
"that multiply to ", showterms([[Math.abs(C),1]]),":
",showfactorpairs(C),"
",
"The sign of the constant term,",showterms([[C,1]]),", is ",(C>0?"positive":"negative"),", so ",showterms([[1,'p']])," and ",
showterms([[1,'q']])," will have ",(C>0?"the same":"opposite")," signs.
",
(C>0?"Since the sign of the "+showterms([[1,v]])+" term is "+(B>0?"positive":"negative")+", "+
showterms([[1,'p']])+" and "+showterms([[1,'q']])+" are both "+(B>0?"positive":"negative")+".
":""),
(fps.length>1?"The only combination that gives a sum of ":"So the pair of signed factors that sum to "),
showterms([[B,1]]),", the coefficient of ",showterms([[1,v]]),", is ",showterms([showsign(a),[Math.abs(a),1]]),
" and ",showterms([showsign(b),[Math.abs(b),1]]),":
",
showterms([[A,v,2],[B,v],[C,1]," = (",[1,v],[a,1],")(",[1,v],[b,1],")"]),
(K>1? "
"+implies+showterms([[K*A,v,2],[K*B,v],[K*C,1]," = ",[K,1],"(",[ka,v],[a,1]," )( ",[kb,v],[b,1],")"]):"")
);
return};
var V=(v=='x'?'y':'x'),P='p',Q='q',R='r',S='s';
if((P+Q+R+S).indexOf(v)!=-1){P='a';Q='b';R='c';S='d'};
document.writeln("To factorize ",showterms([[A,v,2],[B,v],[C,1]]),"
",
"Our reference formula is ",showterms(["(",[1,P+v],[1,Q]," )( ",[1,R+v],[1,S]," ) = ",
[1,P+R+v,2],"+ (",[1,P+S],[1,Q+R],") ",[1,v],[1,Q+S]]),"
",
"Method: find two factors of ",showterms(["coefficient of ",[1,v,2],"×",["constant term",1]," = ",
[1,P+Q+R+S]," = ",[1,P+S]," ×",[1,Q+R]]),"
that sum to the ",showterms([["coefficient of ",v]," = ",[1,P+S],[1,Q+R]]),"
",
"
",
"First find the product of the ",showterms([[1,v,2]]),
" coefficient: ",showterms([[A,1]])," and the constant term: ",showterms([[C,1]]),
" which is ",showterms([[negb(A),1],times,[negb(C),1]," = ",[A*C,1]]),
"
Now find the pairs of factors of this number:",
showfactorpairs(A*C),"
",
(A*C<0
?"but the product we want is "+showterms([[A*C,1]])+" so one of each factor pair must be negative, the other positive."
:" and also the same pairs with both factors being negative."
),
"
Of these choose the (signed) factor pair that sums to ",
" the coefficient of ",showterms([[1,v]]),": ",showterms([[B,1]])," ...",
"The pair to choose is ",showterms([[ka*b,1],", ",[a*kb,1]])," because ",
showterms([[A*C,1]," = ",[negb(ka*b),1],"×",[negb(a*kb),1]])," and ",showterms([[ka*b,1],[a*kb,1]," = ",[B,1]]),
".
We can therefore rewrite our quadratic substituting ",showterms([[ka*b,v],[a*kb,v]])," for ",showterms([[B,v]]),":
",
showterms([[A,v,2],[B,v],[C,1]," = ",[A,v,2],[ka*b,v],[a*kb,v],[C,1]]),
" which enables us to factor the first two and last two terms:
",
showterms([[A,v,2],[B,v],[C,1]," = ",[ka,v],"(",[kb,v],[b,1],")"+(a>0?" + ":""),[a,1],"(",[kb,v],[b,1],")"]),
" and the bracketed term now factors out too:
",
showterms([[A,v,2],[B,v],[C,1]," = (",[ka,v],[a,1],")(",[kb,v],[b,1],")"]),
(K>1?"
"+implies+showterms([[K*A,v,2],[K*B,v],[K*C,1]," = ",[K,1],"(",[ka,v],[a,1]," )( ",[kb,v],[b,1],")"]):"")
);
};
function SOLNfomula(A,B,C,v,ka,a,kb,b){document.writeln("The Formula that gives the two roots of a quadratic is:
",
oneline("If ",showterms([["a ",v,2],["b ",v],["c",1]," = 0"])," then ",showterms([[1,v]," = "]),
Frac("–"+Var('b')+" ± √("+showterms([[1,"b",2 ],[-4,"ac"]],false)+")",
showterms([[2,'a']],"false")+"")),
" and these 2 values of ",showterms([[1,v]])," we will call root1 and root2.
",
"The factors of ",showterms([["a ",v+"2"],["b ",v],["c",1]]),
" will therefore be ( ",v," – root1 ) and ( ",v," – root2 )",
"For this question, we need the roots of ",showterms([[A,v,2],[B,v],[C,1]])," so ",
"a = ",A,", b = ",B," and c = ",C,"
",
"The formula gives the roots ",v," = "+new RAT(-a,ka).toString()," and ",v,
" = "+new RAT(-b,kb).toString()+"
",
"So the factors are (",showterms([[ka,v],[a,1]],false)," ) and ( ",
showterms([[kb,v],[b,1]],false)," )") };