Consider given the polynomial, #10; #10; p( y #10;)=y^2 y+1 #160; #160; #160; #160; #160; #160; #8230; #8230;.(1) #10; #10;We ha ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Graph of Quadratic Expression

Py=y2-y+1 fin...

Question

$P (y) = y^{2} - y + 1$ find $P (0) = ?$

Open in App

Solution

Consider given the polynomial,
$p (y) = y^{2} - y + 1$ …….(1)
We have to find, $p (0)$
Put, $y = 0$ in equation (1)
Now,
$p (0) = 0^{2} - 0 + 1$
$\Rightarrow p (0) = 1$

Suggest Corrections

Similar questions

Find p(0) if p (y) = y² - y + 1

p (0), p (1), p (2)

p (y) = y^{2} - y + 1

Find p(0), p(1) and p(2) for each of the following polynomials:
$1) p (y) = y^{2} - y + 1$
$2) p (t) = 2 + t + 2 t^{2} - t^{3}$ [4 MARKS]

p (y) = y^{2} - y + 1

p (t) = 2 + t + 2 t^{2} - t^{3}

Join BYJU'S Learning Program