Answer:
The required solutions are a. [tex]f(x)=(x-3)(3x+4)[/tex], b. [tex]3\text{ and }-\frac{3}{4}[/tex], c. [tex]3\text{ and }-\frac{3}{4}[/tex].
Step-by-step explanation:
a.
The given quadratic equation is
[tex]f(x)=3x^2-5x-12[/tex]
The middle term can be written as (-9x+4x)
[tex]f(x)=3x^2-9x+4x-12[/tex]
[tex]f(x)=3x(x-3)+4(x-3)[/tex]
[tex]f(x)=(x-3)(3x+4)[/tex]
The intercept form of the given equation by factoring. is,
[tex]f(x)=(x-3)(3x+4)[/tex]
b.
The x-intercepts,zeros, roots and solutions are same things.
[tex]0=(x-3)(3x+4)[/tex]
Equate each factor equal to 0.
[tex]x=3,-\frac{3}{4}[/tex]
Therefore x-intercepts,zeros, roots and solutions of the given equation are [tex]3\text{ and }-\frac{3}{4}[/tex].
c.
The given equation is
[tex]3x^2-5x-12=0[/tex]
Therefore the solution of the equation are [tex]3\text{ and }-\frac{3}{4}[/tex].
