The factoring method which can be considered for such a cubic tetranomial expression is; factor by grouping sum of cubes.
<h3>What factoring method can be considered for the polynomial?</h3>
It follows from the task content that the order of the Polynomial is 3 and the polynomial is a tetranomial as it contains 4 terms.
On this note, since 3x³ is not a perfect cube, it follows that the best factorisation method for such a polynomial is; factor by grouping sum of cubes.
Read more on factorisation;
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F(x) = 2x² - 8x - 10.
This is a parabola open upward (since a>0) with an axis of symmetry = -b/2a:
a) axis of symmetry: x = -(-8)/(2*2) = 8/4 = 2. Then x = 2, which is the x component of the vertex
b) for x = 2, f(x) = f(2) = - 18 (component of y of the vertex)
c) VERTEX(2, - 18)
d) DISCRIMINENT: b² - 4.a.c = 64 - 4*2*(-10) = 144
Answer: To find the area of a rectangle, multiply its height by its width. For a square you only need to find the length of one of the sides (as each side is the same length) and then multiply this by itself to find the area.
It would be 200+9+0.1+0.006.
