Find the [tex]$x$[/tex]-intercepts of [tex]$f(x)=4x^4+12x^3-40x^2$[/tex].

A. [tex]$(0,0), (7,0), (-2,0)$[/tex]
B. [tex]$(0,0), (7,0), (-2,0)$[/tex]
C. [tex]$(0,0), (-2,0), (5,0)$[/tex]
D. [tex]$(0,0), (2,0), (-5,0)$[/tex]



Answer :

To find the x-intercepts of the function \( f(x) = 4x^4 + 12x^3 - 40x^2 \), we need to solve for \( x \) when \( f(x) = 0 \).

1. Set the function equal to zero:
[tex]\[ 4x^4 + 12x^3 - 40x^2 = 0 \][/tex]

2. Factor out the greatest common factor (GCF):
Notice that \( 4x^2 \) is a common factor in each term of the polynomial. Factor out \( 4x^2 \):
[tex]\[ 4x^2(x^2 + 3x - 10) = 0 \][/tex]

3. Solve for \( x \) in \( 4x^2 = 0 \):
[tex]\[ 4x^2 = 0 \implies x^2 = 0 \implies x = 0 \][/tex]
Thus, one x-intercept is \( (0,0) \).

4. Solve the quadratic equation \( x^2 + 3x - 10 = 0 \):
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 3 \), and \( c = -10 \):
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 40}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{49}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 7}{2} \][/tex]

This results in two solutions:
[tex]\[ x = \frac{-3 + 7}{2} = \frac{4}{2} = 2 \][/tex]
[tex]\[ x = \frac{-3 - 7}{2} = \frac{-10}{2} = -5 \][/tex]
Thus, the other two x-intercepts are \( (2,0) \) and \((-5,0) \).

5. Summarize the x-intercepts:
The x-intercepts of the function \( f(x) = 4x^4 + 12x^3 - 40x^2 \) are:
[tex]\[ (0,0), \, (2,0), \, (-5,0) \][/tex]

Therefore, the correct choice is:
[tex]\((0,0), (2,0), (-5,0)\)[/tex].

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