If $x^{2} - 3 x + 2$ is factor of $x^{4} - p x^{2} + q = 0$ then the values of p and q are

p = 5, q = 4

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

p = 4, q = 3

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

p = 6, q = 5

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

p = 5, q = 6

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

Open in App

Solution

The correct option is A $p = 5, q = 4$
$x^{2} - 3 x + 2 = x^{2} - 2 x - x + 2 = x (x - 2) - 1 (x - 2) = (x - 1) (x - 2)$
Factors of $(x^{2} - 3 x + 2)$ are $(x - 1) (x - 2)$ and each factor produces a zero remainder.
$∴ 1 - p + q = 0$ and $16 - 4 p + q = 0$
Solving we get p=5 and q=4.

Suggest Corrections

Similar questions

α, β

x^{2} - p x + q = 0

(α^{2} - β^{2}) (α^{3} - β^{3})

(α^{3} β^{2} + α^{2} β^{3})

(x^{2} - 3 x + 2)

x^{4} - {p x}^{2} + q

x^{2} - 3 x + 2

f (x) = x^{4} - p x^{2} + q

(p, q) =

p^{3} q^{2} - 6 p^{2} q^{5} + p^{4} q^{4} - 8 p q^{6} + 5 p^{3} q

If (7,3), (6,1), (8,2) and (p,q) are the vertices of a parallelogram, taken in order, then find the values of p and q.
p = ___ and q = ___

Join BYJU'S Learning Program