# If  g(x)=x^2+3x+1  then show that g(x+1)-g(x) = 2x+4?

Mar 13, 2017

$\left(i\right) : g \left(x + 1\right) = {x}^{2} + 5 x + 5.$

$\left(i i\right) :$ For, Verification, refer to The Explanation Section.

#### Explanation:

$g \left(x\right) = {x}^{2} + 3 x + 1.$

$\Rightarrow g \left(x + 1\right) = {\left(x + 1\right)}^{2} + 3 \left(x + 1\right) + 1.$

$= {x}^{2} + 2 x + 1 + 3 x + 3 + 1$

$\therefore g \left(x + 1\right) = {x}^{2} + 5 x + 5.$

Next, $g \left(x + 1\right) - g \left(x\right) = {x}^{2} + 5 x + 5 - \left({x}^{2} + 3 x + 1\right)$

$\Rightarrow g \left(x + 1\right) - g \left(x\right) = 2 x + 4.$ Hence, the Verification.

Mar 13, 2017

We have;

$g \left(x\right) = {x}^{2} + 3 x + 1$

And so:

$g \left(x + 1\right) = {\left(x + 1\right)}^{2} + 3 \left(x + 1\right) + 1$
$\text{ } = \left({x}^{2} + 2 x + 1\right) + \left(3 x + 3\right) + 1$
$\text{ } = {x}^{2} + 5 x + 5$

Therefore:

$g \left(x + 1\right) - g \left(x\right) = \left({x}^{2} + 5 x + 2\right) - \left({x}^{2} + 3 x + 1\right)$
$\text{ } = {x}^{2} + 5 x + 5 - {x}^{2} - 3 x - 1$
$\text{ } = 2 x + 4 \setminus \setminus \setminus \setminus \setminus Q E D$