Question

What is the factored form of the equation? Factor completely.

24q7 – 42q4r + 36q3r2 – 63r3

Answer 1

@Callisto

Answer 2

@Lilith

Answer 3

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)

Answer 4

so we have (a*q^3-b*r)(c*q^4+d*r^2)

Answer 5

\[(a*q^3-b*r)(c*q^4+d*r^2)=acq^7-bcq^4r+adq^3r^2-bdr^3\]

Answer 6

ac=24; bc=42; ad=36; bd=63

Answer 7

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)

Answer 8

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;

Answer 9

we can finish the factoring now.
\[(aq^3−br)(cq^4+dr^2)=acq^7−bcq^4r+adq^3r^2−bdr^3\]
\[(12q^3−21r)(2q^4+3r^2)=24q^7−36q^4r+42q^3r^2−63r^3\]

Answer 10

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
\[3(4q^3-21r)(2q^4+3r^2)\]

Answer 11

that should be\[3(4q^3−7r)(2q^4+3r^2)\]