Question

Find the value of x^3-8x^2+14x-7, when x = -1​

Answer 1

Rules Used :-

# Two negatives when multiplied always makes a positive sign.

# A positive and a negative when multiplied always makes negative sign.

# Two negatives always makes positive with a negative sign.

Answer 2

Answer:

$\boxed{ \sf{ \:The \: value \: of{x}^{3} - {8x}^{2} + 14x - 7 \: at \: x = - 1 \: is \: \bf \: - 30 \: }}$

Step-by-step explanation:

Given polynomial is

$\qquad\sf \: {x}^{3} - {8x}^{2} + 14x - 7$

Now, Substitute x = - 1, we get

$\qquad\sf \: = \: {( - 1)}^{3} - {8( - 1)}^{2} + 14( - 1) - 7$

$\qquad\sf \: = \: - 1 - 8(1) - 14 - 7$

$\qquad\sf \: = \: - 1 - 8 - 21$

$\qquad\sf \: = \: - 30$

Hence,

$\sf\implies The \: value \: of{x}^{3} - {8x}^{2} + 14x - 7 \: at \: x = - 1 \: is \: \bf \: - 30$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\:\\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$