Please help this is for rotc it's due at 6

Mathematics

1 answer:

Gnom [1K]2 years ago

5 0

Answer:

UK, Egypt, India, the Philippines, Colombia,

Step-by-step explanation:

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Please help! Thanks in advance.

alukav5142 [94]

In this problem, we need to plug in the given x values for $f(x)=0$ and find a and b.

When we plug in 1, we get: $2 \times {1}^{4} - 5 \times {1}^{3} - 14 \times {1}^{2} + a \times 1 + b = 0$

Simplify:
$2 - 5 - 14 + a + b = 0$
$- 17 + a + b = 0$
$a + b = 17$

We got our first statement about the values of the variables. If we find one more we can find those 2 variables.

We have another given root: 4.

Plug it in:
$2 \times {4}^{4} - 5 \times {4}^{3} - 14 \times {4}^{2} + a \times 4 + b = 0$
$512 - 320 - 224 + 4a + b = 0$
$- 32 + 4a + b = 0$
$4a + b = 32$

Now we have our second one. We can combine them:
$a + b = 17 \\ 4a + b = 32$

I use elimination method which is easier here.

Multiply the top equation by -1: $- a - b = - 17 \\ 4a + b = 32$

Add them up:
$3a = 15$

Simplify:
$a = 5$

Now we have a, we can plug in one of those equations to find b:
$5 + b = 17$
$b = 17 - 5$
$b = 12$

So, the answers are $a=5$ and $b=12$ .