Answer:
3(g²+3g-2)
Step-by-step explanation:
Factor the common term:
(3g2+9g−6)=(3(g2+3g−2))
To factor the quadratic function g2+3g−2g2+3g−2, we should solve the corresponding quadratic equation g2+3g−2=0g2+3g−2=0.
Indeed, if g1g1 and g2g2 are the roots of the quadratic equation ag2+bg+c=0ag2+bg+c=0, then ag2+bg+c=a(g−g1)(g−g2)ag2+bg+c=a(g−g1)(g−g2).
Solve the quadratic equation g2+3g−2=0g2+3g−2=0.
The roots are g1=−32+17−−√2g1=−32+172, g2=−32+17−−√2g2=−32+172 (use the quadratic equation calculator to see the steps).
Since the roots are irrational, we do not factor further. Therefore, we leave g2+3g−2g2+3g−2 as it is.
Rewrite:
3(g2+3g−2)=3g2+9g−6
Therefore, 3g2+9g−6=3g2+9g−63g2+9g−6=3g2+9g−6.
Next steps are not available, but the factorization is
Thus, 3g2+9g−6=3(g2+3g−2)3g2+9g−6=3(g2+3g−2).
Answer: 3g2+9g−6=3(g2+3g−2)3g2+9g−6=3(g2+3g−2).
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Answers & Comments
Answer:
3(g²+3g-2)
Step-by-step explanation:
Factor the common term:
(3g2+9g−6)=(3(g2+3g−2))
(3g2+9g−6)=(3(g2+3g−2))
To factor the quadratic function g2+3g−2g2+3g−2, we should solve the corresponding quadratic equation g2+3g−2=0g2+3g−2=0.
Indeed, if g1g1 and g2g2 are the roots of the quadratic equation ag2+bg+c=0ag2+bg+c=0, then ag2+bg+c=a(g−g1)(g−g2)ag2+bg+c=a(g−g1)(g−g2).
Solve the quadratic equation g2+3g−2=0g2+3g−2=0.
The roots are g1=−32+17−−√2g1=−32+172, g2=−32+17−−√2g2=−32+172 (use the quadratic equation calculator to see the steps).
Since the roots are irrational, we do not factor further. Therefore, we leave g2+3g−2g2+3g−2 as it is.
Rewrite:
3(g2+3g−2)=3g2+9g−6
3(g2+3g−2)=3g2+9g−6
Therefore, 3g2+9g−6=3g2+9g−63g2+9g−6=3g2+9g−6.
Next steps are not available, but the factorization is
Thus, 3g2+9g−6=3(g2+3g−2)3g2+9g−6=3(g2+3g−2).
Answer: 3g2+9g−6=3(g2+3g−2)3g2+9g−6=3(g2+3g−2).