Question

optional challenge:
a) write an equation that defines m(x) as a trinomial where m(x) = (3x - 1)(3 - x) + 4x² + 19.
b) solve for x by factoring when m(x) = 0.
c) assess the validity of your answers.

Explanation:

Step1: Expand the binomial product

$(3x - 1)(3 - x) = 3x\cdot3 - 3x\cdot x -1\cdot3 +1\cdot x = 9x - 3x^2 - 3 + x$

Step2: Combine like terms with $4x^2+19$

$m(x) = -3x^2 + 9x + x - 3 + 4x^2 + 19 = (-3x^2+4x^2)+(10x)+(-3+19)$
$m(x) = x^2 + 10x + 16$

Step3: Factor the trinomial for $m(x)=0$

Find two numbers that multiply to $16$ and add to $10$: $2$ and $8$.
$x^2 + 10x + 16 = (x+2)(x+8) = 0$

Step4: Solve for $x$ using zero product rule

If $(x+2)=0$, then $x=-2$; if $(x+8)=0$, then $x=-8$

Step5: Substitute $x=-2$ into original $m(x)$

$m(-2)=(3(-2)-1)(3-(-2))+4(-2)^2+19=(-7)(5)+16+19=-35+35=0$

Step6: Substitute $x=-8$ into original $m(x)$

$m(-8)=(3(-8)-1)(3-(-8))+4(-8)^2+19=(-25)(11)+256+19=-275+275=0$

Answer:

a) $m(x) = x^2 + 10x + 16$
b) $x=-2$ and $x=-8$
c) Both $x=-2$ and $x=-8$ are valid, as substituting them back into the original $m(x)$ results in $0$, satisfying $m(x)=0$.