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2 years ago

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Sena determines the remainder of 4x32−14x16+2x3−8x−1, using the remainder theorem. How does she proceed to the correct answer? D

rag a value into each box to correctly complete the statements.

1 answer:

Flauer [41]2 years ago

4 0

Answer: x = 1 and -16

Explanation:

(4x^32 - 14x^16 + 2x^3 - 8)/x - 1

x - 1 = 0
x = 1

f(x) = 4x^32 - 14x^16 + 2x^3 - 8
f(1) = 4(1)^32 - 14(1)^16 + 2(1)^3 - 8
f(1) = 4 - 14 + 2 - 8
f(1) = -10 - 6
f(1) = -16

Therefore, the remainder is -16

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How many solutions does the equation have, x1 + x2 + x3 = 10 , where x1 , x2, and x3 are non-negative integers?

MAVERICK [17]

Non-negative integers are positive integers or zero.

1. When $x_1=0,$ then there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=10;$
$x_2=1,\ x_3=9;$
$x_2=2,\ x_3=8;$
$x_2=3,\ x_3=7;$
$x_2=4,\ x_3=6;$
$x_2=5,\ x_3=5;$
$x_2=6,\ x_3=4;$
$x_2=7,\ x_3=3;$
$x_2=8,\ x_3=2;$
$x_2=9,\ x_3=1;$
$x_2=10,\ x_3=0.$

In total 11 solutions for $x_1=0.$

2. For $x_1=1,$ there are such possible cases for $x_2$ and $x_3:$

$x_2=0,\ x_3=9;$
$x_2=1,\ x_3=8;$
$x_2=2,\ x_3=7;$
$x_2=3,\ x_3=6;$
$x_2=4,\ x_3=5;$
$x_2=5,\ x_3=4;$
$x_2=6,\ x_3=3;$
$x_2=7,\ x_3=2;$
$x_2=8,\ x_3=1;$
$x_2=9,\ x_3=0.$

In total 10 solutions for $x_1=1.$

3. This process gives you

for $x_1=2$ - 9 solutions;
for $x_1=3$ - 8 solutions;
for $x_1=4$ - 7 solutions;
for $x_1=5$ - 6 solutions;
for $x_1=6$ - 5 solutions;
for $x_1=7$ - 4 solutions;
for $x_1=8$ - 3 solutions;
for $x_1=9$ - 2 solutions;
for $x_1=10$ - 1 solution.

4. Add all numbers of solutions:

$11+10+9+8+7+6+5+4+3+2+1=66.$

Answer: there are 66 possible solutions (with non-negative integer variables)

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