Question

I will mark brainliest if you give correct answer on time

Answer 1

Step-by-step explanation:

3a²-7a−6

Factor the expression by grouping. First, the expression needs to be rewritten as 3a

2

+pa+qa−6. To find p and q, set up a system to be solved.

p+q=−7

pq=3(−6)=−18

Since pq is negative, p and q have the opposite signs. Since p+q is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product −18.

1,−18

2,−9

3,−6

Calculate the sum for each pair.

1−18=−17

2−9=−7

3−6=−3

The solution is the pair that gives sum −7.

p=−9

q=2

Rewrite 3a

2−7a−6 as (3a

2−9a)+(2a−6).

(3a 2−9a)+(2a−6)

Factor out 3a in the first and 2 in the second group.

3a(a−3)+2(a−3)

Factor out common term a−3 by using distributive property.

(a−3)(3a+2)

Answer 2

Answer:

$\purple{ \boxed{ \bold{ { \bigg(9a - \frac{21}{2} \bigg)}^{2} - \bigg(\frac{33}{2} \bigg)^{2} }}}$

Step-by-step explanation:

$3a^2- 7a- 6 \\ \\ \implies \: take \: 3 \: as \: common \\ = 3 \bigg \{ {a}^{2} - \frac{7}{3} a - 2 \bigg \} \\ \\ = 3 \bigg \{{a}^{2} - 2 \times \frac{7}{6} a + \bigg( \frac{7}{6} \bigg)^{2} - \bigg( \frac{7}{6} \bigg)^{2} - 2 \bigg \} \\ \\ = 3 \bigg \{ { \bigg(a - \frac{7}{6} \bigg)}^{2} - \frac{49}{36} - 2 \bigg \} \\ \\ = 3 \bigg \{ { \bigg(a - \frac{7}{6} \bigg)}^{2} - \bigg(\frac{49 + 72}{36} \bigg) \bigg \} \\ \\ = 3 \bigg \{ { \bigg(a - \frac{7}{6} \bigg)}^{2} - \bigg(\frac{121}{36} \bigg) \bigg \} \\ \\ = 3 \bigg \{ { \bigg(a - \frac{7}{6} \bigg)}^{2} - \bigg(\frac{11}{6} \bigg)^{2} \bigg \} \\ \\ = { \boxed{ \bold{3 { \bigg(a - \frac{7}{6} \bigg)}^{2} - 3\bigg(\frac{11}{6} \bigg)^{2} }}} \\ \\ = { \boxed{ \bold{ { \bigg(9a - \frac{9\times 7}{6} \bigg)}^{2} - \bigg(\frac{9\times 11}{6} \bigg)^{2} }}}\\ \\ = \purple{ \boxed{ \bold{ { \bigg(9a - \frac{21}{2} \bigg)}^{2} - \bigg(\frac{33}{2} \bigg)^{2} }}}$