HELP ASAP!!! IM BEING TIMED. Use synthetic division to test one potential root. Enter the numbers that complete the division problem.

HELP ASAP IM BEING TIMED Use synthetic division to test one potential root Enter the numbers that complete the division problem class=

Respuesta :

Step-by-step explanation:

1 * (-5) = -5     a = -5

6 + (-5) = 1      b = 1

1 * (-5 ) = -5     c = -5

-7 + -5 = -12     d = -12

The divisor of the polynomial in the synthetic long division is linear factor

The correct values are;

  • a = -5
  • c = -5
  • b = 1
  • d = -12

Reason:

[tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &a & c & 60\end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1 \ \ b \hspace {0.25 cm} d \hspace {0.25 cm} 0[/tex]

The divisor in the division is (x - (-5)) = (x + 5)

By synthetic long division, we have;

Carry down the 1 representing the leading coefficient

  • [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ & & & \end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1[/tex]

Multiply the 1 brought down by the zero value of x which is -5, and take the result into the second line of the division symbol

-5 × 1 = -5

  • [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &-5 & & \end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1[/tex]

Add the coefficient 6 to -5, and bring down the result

  • [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &-5 & & \end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1 \ \ \ 1[/tex]

Repeat the above steps again to get;

-5 × 1 = -5

  • [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &-5 & -5 & \end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1 \ \ \ 1 \hspace {0.3 cm} -12[/tex]

-5 × (-12) = 60

  • [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &-5 & -5 & 60\end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1 \ \ \ 1 \hspace {0.25 cm} -12 \hspace {0.25 cm} 0[/tex]

By comparison to the given synthetic long division, [tex]-5 \underline{\left \lfloor{ \begin{matrix}1 & 6 & -7 &-60 \\ &a & c & 60\end{matrix}}} \underset \hspace {} \hspace {0.6 cm} 1 \ \ b \hspace {0.25 cm} d \hspace {0.25 cm} 0[/tex], we have;

a = -5, c = -5, b = 1, and d = -12

Learn more about synthetic long division here:

https://brainly.com/question/14261063

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