# P(x) is a polynomial function. If P(x^2) = (a-b+2)x^3 - 2x^2 + (2a+b+7)x - 20 , what is P(a+b) ?

Jun 20, 2016

$P \left(x\right)$ cannot be a polynomial because $P \left({x}^{2}\right)$ would certainly be an even function.

#### Explanation:

If $P \left(x\right) = {\sum}_{i = 0}^{n} {a}_{i} {x}^{i}$ then

$P \left({x}^{2}\right) = {\sum}_{i = 0}^{n} {a}_{i} {\left({x}^{2}\right)}^{i} = {\sum}_{i = 0}^{n} {a}_{i} {x}^{2 i}$

The proposition is not feasible once defined $P \left(x\right)$ as a polynomial.

Jun 20, 2016

$P \left(a + b\right) = - 12$

#### Explanation:

$P \left({x}^{2}\right) = \left(a - b + 2\right) {x}^{3} - 2 {x}^{2} + \left(2 a + b + 7\right) x - 20$

Since $P \left(x\right)$ is a polynomial function, any powers of $x$ in $P \left({x}^{2}\right)$ must be even, not odd.

So we require $\left(a - b + 2\right) = 0$ and $\left(2 a + b + 7\right) = 0$

Adding these two equations together, we get: $3 a + 9 = 0$, hence $a = - 3$ and $b = - 1$

$P \left({x}^{2}\right) = - 2 {x}^{2} - 20$

So:

$P \left(x\right) = - 2 x - 20$

Hence:

$P \left(a + b\right)$

$= P \left(\left(- 3\right) + \left(- 1\right)\right)$

$= P \left(- 4\right)$

$= - 2 \left(- 4\right) - 20$

$= 8 - 20$

$= - 12$