Question

7. Simplify the following expressions
c. (6 + 11v)^2 - (6u - 11v)^2​

Answer 1

Answer:

2v2−11v−6

Step-by-step explanation:

Step by Step Solution:

More Icon

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "v2" was replaced by "v^2".

STEP

1

:

Equation at the end of step 1

(2v2 - 11v) - 6

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 2v2-11v-6

The first term is, 2v2 its coefficient is 2 .

The middle term is, -11v its coefficient is -11 .

The last term, "the constant", is -6

Step-1 : Multiply the coefficient of the first term by the constant 2 • -6 = -12

Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is -11 .

-12 + 1 = -11 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -12 and 1

2v2 - 12v + 1v - 6

Step-4 : Add up the first 2 terms, pulling out like factors :

2v • (v-6)

Add up the last 2 terms, pulling out common factors :

1 • (v-6)

Step-5 : Add up the four terms of step 4 :

(2v+1) • (v-6)

Which is the desired factorization

Final result :

(v - 6) • (2v + 1)

hope it will help you

mark me brainliest

Answer 2

Step-by-step explanation:

$c) \: ( {6 + 11v)}^{2} - ( {6u - 11v)}^{2}$

$= 36 + 121 {v}^{2} + 132v - (36 {u}^{2} + 121 {v}^{2} - 132uv)$

$= 36 + 121 {v}^{2} + 132v - 36 {u}^{2} - 121 {v}^{2} + 132uv$

$= 36 + 132v - 36 {u}^{2} + 132uv$

$- 36 {u}^{2} + 132v + 132uv + 36 = 0$

$36 {u}^{2} - 132v - 132uv - 36 = 0$

$dividing \: each \: term \: by \: 36$

${u}^{2} - 22v - 22uv - 1 = 0$