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

`3x^3-39x-36`

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

`{ \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