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In the synthetic-division problem shown below, what number belongs in the place of the question mark?

[tex]\[
\begin{array}{r|rrrrr}
2 & 1 & -2 & -8 & 5 \\
& & & & & \\
\hline
& & & & &
\end{array}
\][/tex]

[tex]\[
\frac{2 \quad ?}{10}
\][/tex]


Sagot :

Sure, let's tackle this synthetic-division problem step-by-step!

Given the polynomial [tex]\( 2x^4 + x^3 - 2x^2 - 8x + 5 \)[/tex] and dividing it by [tex]\( x - 2 \)[/tex]:

1. Set up the synthetic division: Write down the coefficients of the polynomial:
[tex]\[ \begin{array}{ccccc} 2 & 1 & -2 & -8 & 5 \\ \end{array} \][/tex]

2. Divide by [tex]\( x - 2 \)[/tex]: The divisor here is 2.

3. Carry down the leading coefficient (which is 2) to the bottom row:
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2 & & & & & 2 \\ \end{array} \][/tex]

4. Multiply the 2 (bottom row) by the divisor (2) and write the result underneath the next coefficient (1):
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & - & 4 & & & 2 \\ \end{array} \][/tex]
Here, [tex]\( 2 \times 2 = 4 \)[/tex].

5. Add the numbers in the second column: [tex]\( 1 + 4 = 5 \)[/tex]:
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & & & 2 \\ & 2 & 5 & & & 2 \\ \end{array} \][/tex]

6. Multiply the recent bottom row value by the divisor (2) and place the result in the next column:
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & 10 & & 2 \\ & 2 & 5 & 8 & & 2 \\ \end{array} \][/tex]
Here, [tex]\( 5 \times 2 = 10 \)[/tex].

7. Add the numbers in the third column: [tex]\( -2 + 10 = 8 \)[/tex]:
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & 10 & 16 & 2 \\ & 2 & 5 & 8 & 8 & 2 \\ \end{array} \][/tex]

8. Next, [tex]\( 8 \times 2 = 16 \)[/tex] (place in the next column):
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & 10 & 16 & & 2 \\ & 2 & 5 & 8 & 8 & 21 & 2 \\ \end{array} \][/tex]

9. Add the numbers in the fourth column: [tex]\( -8 + 16 = 8 \)[/tex]:
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & 10 & 16 & & 2 \\ & 2 & 5 & 8 & 16 & 2 \\ \end{array} \][/tex]

10. Finally, multiply [tex]\( 8 \times 2 = 16 \)[/tex] (place in the next column):
[tex]\[ \begin{array}{ccccc} & 2 & 1 & -2 & -8 & 5 \\ 2: & & 4 & 10 & 16 & 21 & 2 \\ \end{array} \][/tex]

11. Add the numbers in the fifth column: [tex]\( 5 + 16 = 21 \)[/tex].

The final coefficients after synthetic division are:
[tex]\[ 2, 5, 8, 8, 21 \][/tex]

So, the numbers obtained from the synthetic division are [tex]\(2, 5, 8, 8, 21\)[/tex].

Thus, the number that belongs in the place of the question mark is [tex]\(21\)[/tex].
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