Question

f(x)=x2+bx+c that has zeros 8 and 11.

Answer 1

Here you go .hopefully this helps you

Answer 2

Answer:

the polynomial function$F(x) = x^2 - 19x + 88$has zeros of 8 and 11.

Step-by-step explanation:

If a polynomial function $F(x) = x^2 + bx + c$ has zeros of 8 and 11, that means that the equation F(x) = 0 has solutions of x = 8 and x = 11. We can use this information to find the values of b and c.

We know that when x = 8, $F(8) = 8^2 + b8 + c = 0$ and when x = 11, $F(11) = 11^2 + b11 + c = 0.$

So,

$(8^2 + b*8 + c) = 0$

$64 + 8b + c = 0$

and

$(11^2 + b*11 + c) = 0$

$121 + 11b + c = 0$

Now we can solve these two equations simultaneously to find the values of b and c.

From the first equation:

$c = -64 - 8b$

Substitute this value of c in the second equation:

$121 + 11b -64 - 8b = 0$

$57 + 3b = 0$

$3b = -57$

$b = -19$

Now we can substitute the value of b in the first equation to find c

$c = -64 - 8(-19)$

$c = -64 + 152$

$c = 88$

So the polynomial function$F(x) = x^2 - 19x + 88$has zeros of 8 and 11.

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