Solving the expression
[tex]4g^2-g[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c[/tex]
[tex]g^2=gg[/tex]
So,
[tex]4gg-g[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}g[/tex]
[tex]g\left(4g-1\right)[/tex]
So,
[tex]4g^2-g:\quad g\left(4g-1\right)[/tex]
Therefore, the statement [tex]4g^2-g=g^2\left(4-g\right)[/tex] is NOT CORRECT.
Solving the expression
[tex]35g^5-25g^2[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c[/tex]
[tex]g^5=g^3g^2[/tex]
So,
[tex]35g^3g^2-25g^2[/tex]
[tex]\mathrm{Rewrite\:}25\mathrm{\:as\:}5\cdot \:5[/tex]
[tex]\mathrm{Rewrite\:}35\mathrm{\:as\:}5\cdot \:7[/tex]
[tex]5\cdot \:7g^3g^2-5\cdot \:5g^2[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}5g^2[/tex]
[tex]5g^2\left(7g^3-5\right)[/tex]
So,
[tex]35g^5-25g^2=\:5g^2\left(7g^3-5\right)[/tex]
Therefore, the statement [tex]35g^5-25g^2=\:5g^2\left(7g^3-5\right)[/tex] is CORRECT.
Solving the expression
So,
Therefore, the statement [tex]24g^4+18g^2=\:6g^2\left(4g^2+3g\right)[/tex] is CORRECT.
Solving the expression
[tex]9g^3+12[/tex]
[tex]\mathrm{Rewrite\:}12\mathrm{\:as\:}3\cdot \:4[/tex]
[tex]\mathrm{Rewrite\:}9\mathrm{\:as\:}3\cdot \:3[/tex]
[tex]3\cdot \:3g^3+3\cdot \:4[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}3[/tex]
[tex]3\left(3g^3+4\right)[/tex]
So,
[tex]9g^3+12=\:3\left(3g^3+4\right)[/tex]
Therefore, the statement [tex]9g^3+12=\:3\left(3g^3+4\right)[/tex] is CORRECT.
Solving first expression
Considering the expression
[tex]30a^6-24a^2[/tex]
[tex]30a^4a^2-24a^2[/tex]
[tex]\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4[/tex]
[tex]\mathrm{Rewrite\:}30\mathrm{\:as\:}6\cdot \:5[/tex]
[tex]6\cdot \:5a^4a^2-6\cdot \:4a^2[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}3a^2[/tex]
[tex]3a^2\left(10a^4-8\right)[/tex]
Thus, the expression [tex]30a^6-24a^2=3a^2\left(10a^4-8\right)\:[/tex] is completely factored.
Solving second expression
Considering the expression
[tex]12a^3-8a[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c[/tex]
[tex]a^3=a^2a[/tex]
So,
[tex]12a^2a-8a[/tex]
[tex]\mathrm{Rewrite\:}8\mathrm{\:as\:}4\cdot \:2[/tex]
[tex]\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3[/tex]
[tex]4\cdot \:3a^2a-4\cdot \:2a[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}4[/tex]
[tex]4\left(3a^3-2a\right)[/tex]
Thus, the expression [tex]12a^3-8a=\:4\left(3a^3-2a\right)[/tex] is completely factored.
Solving third expression
[tex]16a^5-20a^3[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c[/tex]
[tex]a^5=a^2a^3[/tex]
So,
[tex]16a^2a^3-20a^3[/tex]
[tex]\mathrm{Rewrite\:}20\mathrm{\:as\:}4\cdot \:5[/tex]
[tex]\mathrm{Rewrite\:}16\mathrm{\:as\:}4\cdot \:4[/tex]
[tex]4\cdot \:4a^2a^3-4\cdot \:5a^3[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}4a^3[/tex]
[tex]4a^3\left(4a^2-5\right)[/tex]
Thus, the expression [tex]16a^5-20a^3\:=4a^3\left(4a^2-5\right)[/tex] is completely factored.
Solving fourth expression
[tex]24a^4+18[/tex]
[tex]\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3[/tex]
[tex]\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4[/tex]
[tex]6\cdot \:4a^4+6\cdot \:3[/tex]
[tex]\mathrm{Factor\:out\:common\:term\:}6[/tex]
[tex]6\left(4a^4+3\right)[/tex]
Thus, the expression [tex]24a^4+18=6\left(4a^4+3\right)[/tex] is completely factored.
Keywords: expression, factoring
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Answer:
its 1 and 3 i think! :)
Step-by-step explanation:
