Question

TIMED PLEASE HURRY HELP WITH 2 EASY QUESTIONS

Answer 1

                        Question # 1

Which statements are true?

Analyzing and solving the first statement:

• $4g^2-g=g^2\left(4-g\right)$

Solving the expression

$4g^2-g$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^2=gg$

So,

$4gg-g$

$\mathrm{Factor\:out\:common\:term\:}g$

$g\left(4g-1\right)$

So,

$4g^2-g:\quad g\left(4g-1\right)$

Therefore, the statement $4g^2-g=g^2\left(4-g\right)$ is NOT CORRECT.

Analyzing and solving the second statement:

• $35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Solving the expression

$35g^5-25g^2$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$g^5=g^3g^2$

So,

$35g^3g^2-25g^2$

$\mathrm{Rewrite\:}25\mathrm{\:as\:}5\cdot \:5$

$\mathrm{Rewrite\:}35\mathrm{\:as\:}5\cdot \:7$

$5\cdot \:7g^3g^2-5\cdot \:5g^2$

$\mathrm{Factor\:out\:common\:term\:}5g^2$

$5g^2\left(7g^3-5\right)$

So,

$35g^5-25g^2=\:5g^2\left(7g^3-5\right)$

Therefore, the statement $35g^5-25g^2=\:5g^2\left(7g^3-5\right)$ is CORRECT.

Analyzing and solving the third statement:

• $24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$

Solving the expression

$24g^4+18g^2$

$24g^2g^2+18g^2$

$\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$6\cdot \:4g^2g^2+6\cdot \:3g^2$

$\mathrm{Factor\:out\:common\:term\:}6g^2$

$6g^2\left(4g^2+3\right)$

So,

$24g^4+18g^2=6g^2\left(4g^2+3\right)$

Therefore, the statement $24g^4+18g^2=\:6g^2\left(4g^2+3g\right)$  is CORRECT.

Analyzing and solving the fourth statement:

• $9g^3+12=\:3\left(3g^3+4\right)$

Solving the expression

$9g^3+12$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}3\cdot \:4$

$\mathrm{Rewrite\:}9\mathrm{\:as\:}3\cdot \:3$

$3\cdot \:3g^3+3\cdot \:4$

$\mathrm{Factor\:out\:common\:term\:}3$

$3\left(3g^3+4\right)$

So,

$9g^3+12=\:3\left(3g^3+4\right)$

Therefore, the statement $9g^3+12=\:3\left(3g^3+4\right)$ is CORRECT.

                        Question # 2

Which expressions are completely factored?

Solving first expression

Considering the expression

• $30a^6-24a^2$

$30a^6-24a^2$

$30a^4a^2-24a^2$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$\mathrm{Rewrite\:}30\mathrm{\:as\:}6\cdot \:5$

$6\cdot \:5a^4a^2-6\cdot \:4a^2$

$\mathrm{Factor\:out\:common\:term\:}3a^2$

$3a^2\left(10a^4-8\right)$

Thus, the expression $30a^6-24a^2=3a^2\left(10a^4-8\right)\:$ is completely factored.

Solving second expression

Considering the expression

• $12a^3-8a$

$12a^3-8a$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^3=a^2a$

So,

$12a^2a-8a$

$\mathrm{Rewrite\:}8\mathrm{\:as\:}4\cdot \:2$

$\mathrm{Rewrite\:}12\mathrm{\:as\:}4\cdot \:3$

$4\cdot \:3a^2a-4\cdot \:2a$

$\mathrm{Factor\:out\:common\:term\:}4$

$4\left(3a^3-2a\right)$

Thus, the expression $12a^3-8a=\:4\left(3a^3-2a\right)$ is completely factored.

Solving third expression

• $16a^5-20a^3\:\:\:\:\:\:\:\:\:\:\:\:$

$16a^5-20a^3$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c$

$a^5=a^2a^3$

So,

$16a^2a^3-20a^3$

$\mathrm{Rewrite\:}20\mathrm{\:as\:}4\cdot \:5$

$\mathrm{Rewrite\:}16\mathrm{\:as\:}4\cdot \:4$

$4\cdot \:4a^2a^3-4\cdot \:5a^3$

$\mathrm{Factor\:out\:common\:term\:}4a^3$

$4a^3\left(4a^2-5\right)$

Thus, the expression $16a^5-20a^3\:=4a^3\left(4a^2-5\right)$ is completely factored.

Solving fourth expression

• $24a^4+18$

$24a^4+18$

$\mathrm{Rewrite\:}18\mathrm{\:as\:}6\cdot \:3$

$\mathrm{Rewrite\:}24\mathrm{\:as\:}6\cdot \:4$

$6\cdot \:4a^4+6\cdot \:3$

$\mathrm{Factor\:out\:common\:term\:}6$

$6\left(4a^4+3\right)$

Thus, the expression $24a^4+18=6\left(4a^4+3\right)$ is completely factored.

Keywords: expression, factoring

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