Question

h(x) = 3(x+1)(x+3)(x-4) for each polynomial function rewrite in standard form and state its degree and constant

Answer 1

Answer:

$3x^3-39x-36$

Step-by-step explanation:

$\mathrm{Expand\:}\left(x+1\right)\left(x+3\right) by applying the FOIL method: \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

$\left(x+1\right)\left(x+3\right)=x^2+x\cdot \:3+1\cdot \:x+1\cdot \:3$

$=x^2+4x+3$

$\mathrm Thus 3\left(x+1\right)\left(x+3\right)\left(x-4\right)\\\\=3\left(x^2+4x+3\right)\left(x-4\right)$

$= (3x^2 + 12x + 9)(x - 4)$

$= x(3x^2 + 12x + 9) -4 (3x^2 + 12x + 9)\\\\= 3x^3 + 12x^2 + 3x - 12x^2 - 48x - 36)\\\\= 3x^3 + (12x^2 - 12x^2) + (9x - 48x) -36\\\\= 3x^3-39x-36$

Answer 2

Answer:

${ \tt{h(x) = 3(x + 1)(x + 3)(x - 4)}} \\ \\ { \tt{h(x) = 3 \{( {x}^{2} + 4x + 3)(x - 4) \} }} \\ \\ { \tt{h(x) = 3 \{ {(x}^{3} - {4x}^{2} + {4x}^{2} - 8x + 3x - 12) \} }} \\ \\ { \tt{h(x) = 3( {x}^{3} - 5x - 12) }} \\ \\ { \tt{h(x) = 3 {x}^{3} - 15x - 36 }}$

Third degree polynomial with a constant of -36