2 years ago

Help me out here guys

1 answer:

3 0

D.

f(x) can be written as (x+2)(x-2)(x-1)
by using difference of two squares to expand x^2 - 4 whch yields 3 x intercepts

similarly, k(x) can be written as
x(x+5)(x-5) which also yields 3
x intercepts (0,-5, and 5)

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Yuri is not correct.

Step-by-step explanation:

Given expression is q(x) = 6x³ + 19x² - 15x - 28

If 'a' is a root of the given function, then by substituting x = a in the expression, q(a) = 0

Similarly, for x = $\frac{3}{4}$ ,

$q(\frac{3}{4})=6(\frac{3}{4})^3+19(\frac{3}{4})^2-15(\frac{3}{4})-28$

= $6(\frac{27}{64})+19(\frac{9}{16})-15(\frac{3}{4})-28$

= $(\frac{162}{64})+(\frac{171}{16})-(\frac{45}{4})-28$

= $(\frac{162}{64})+(\frac{684}{64})-(\frac{720}{64})-\frac{1792}{64}$

= $-\frac{1666}{64}$

= $-\frac{833}{32}$ ≠ 0

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