Find the absolute value of the polynomial $5 x - 4 x^{2} + 3$ at $x = - 1$

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Solution

Let $f (x) = 5 x - 4 x^{2} + 3$
value of f(x) at $x = - 1 = f (- 1)$
$∴ f (- 1) = 5 (- 1) - 4 (- 1)^{2} + 3 = - 5 - 4 + 3 = - 9 + 3 = - 6$
$∴$ The absolute value is $6$

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Similar questions

5 x - 4 x^{2} + 3

x = - 1

5 x + 4 x^{2} + 3

x = - 1

5 x - 4 x^{2} + 3

x = 2

4 x^{2} - 5 x + 3

x = \frac{1}{2}

Find the value of the polynomial $5 x - 4 x^{2} + 3$ at
(i) $x = 0$
(ii) $x = - 1$
(iii) $x = 2$

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