Find the value of $3 x^{2} - 4 x^{2} + 5 x - 6,$ when $x = - 1$

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Solution

Substituting $x = - 1$ in $3 x^{2} - 4 x^{2} + 5 x - 6$

$3 x^{2} - 4 x^{2} + 5 x - 6$
$= 3 {(- 1)}^{3} - 4 {(- 1)}^{2} + 5 (- 1) - 6$
$[∵ odd multiple of negative number is negative number]$
$= - 3 - 4 - 5 - 6$
$= - 18$

Therefore,the value of $3 x^{2} - 4 x^{2} + 5 x - 6$ when $x = - 1$ is $- 18$ .

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Similar questions

3 x^{3} - 4 x^{2} + 5 x - 6

x = 1

4 x^{2} - 5 x + 3

x = \frac{1}{2}

4 x^{2} - 5 x + 3

x = - 1

3 x^{3} - 4 x^{2} + 5 x - 6

x = (- 1)

The value of the polynomial $5 x - 4 x^{2} + 3$ , when $x = - 1$ is
A) $- 6$
B) $6$
C) $2$
D) $- 2$

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