If $p (x) = | x^{2} + 2 x + 1 |$ , find the value of $p (- 4)$ .

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-9

No worries! We‘ve got your back. Try BYJU‘S free classes today!

-5

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 9
Given, $p (x) = x^{2} + 2 x + 1$
The value of $p (- 4)$ is
$\Rightarrow p (- 4) = (- 4)^{2} + 2 \times - 4 + 1$
$\Rightarrow p (- 4) = 16 - 8 + 1$
$\Rightarrow p (- 4) = 9$

Suggest Corrections

Similar questions

p (x) = x^{2} - 2 x + 5

x = 2 {log}_{2}^{8}

P (x) = x^{2} - 2 x + 5

x = - 1

x = 5

x = 9 + 4 \sqrt{5}

x^{2} + \frac{1}{x^{2}}

\sqrt{5}

x^{2} + \frac{1}{x^{2}}

p (x) = x^{4} - {2 x}^{3} + {3 x}^{2} - p x + 3 p - 7

p (2) = 5

Join BYJU'S Learning Program