# The sum of two polynomials is 10a^2b^2-9a^2b+6ab^2-4ab+2. If one addend is -5a^2b^2+12a^2b-5, what is the other addend?

Aug 27, 2017

See a solution process below:

#### Explanation:

Let's call the second addend: $x$

We can then write:

$x + \left(- 5 {a}^{2} {b}^{2} + 12 {a}^{2} b - 5\right) = 10 {a}^{2} {b}^{2} - 9 {a}^{2} b + 6 a {b}^{2} - 4 a b + 2$

To find the second addend we can solve for $x$:

$x + \left(- 5 {a}^{2} {b}^{2} + 12 {a}^{b} - 5\right) - \left(- 5 {a}^{2} {b}^{2} + 12 {a}^{2} b - 5\right) =$
$10 {a}^{2} {b}^{2} - 9 {a}^{2} b + 6 a {b}^{2} - 4 a b + 2 - \left(- 5 {a}^{2} {b}^{2} + 12 {a}^{2} b - 5\right)$

$x + 0 = 10 {a}^{2} {b}^{2} - 9 {a}^{2} b + 6 a {b}^{2} - 4 a b + 2 + 5 {a}^{2} {b}^{2} - 12 {a}^{2} b + 5$

$x = 10 {a}^{2} {b}^{2} - 9 {a}^{2} b + 6 a {b}^{2} - 4 a b + 2 + 5 {a}^{2} {b}^{2} - 12 {a}^{2} b + 5$

We can now group and combine like terms:

$x = 10 {a}^{2} {b}^{2} + 5 {a}^{2} {b}^{2} - 9 {a}^{2} b - 12 {a}^{2} b + 6 a {b}^{2} - 4 a b + 2 + 5$

$x = \left(10 + 5\right) {a}^{2} {b}^{2} + \left(- 9 - 12\right) {a}^{2} b + 6 a {b}^{2} - 4 a b + \left(2 + 5\right)$

$x = 15 {a}^{2} {b}^{2} + \left(- 21\right) {a}^{2} b + 6 a {b}^{2} - 4 a b + 7$

$x = 15 {a}^{2} {b}^{2} - 21 {a}^{2} b + 6 a {b}^{2} - 4 a b + 7$