The factors of ${(2 x^{2} - x - 6)}^{2} - (2 x^{2} - 9 x + 10)^{2}$ are:

8 (x - 2) {(2 x - 1)}^{2}

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

16 {(x - 2)}^{2} (2 x - 1)

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

16 (x - 2) (2 x - 1)

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

16 {(x - 2)}^{2} {(2 x - 1)}^{2}

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

Open in App

Solution

The correct option is B $16 {(x - 2)}^{2} (2 x - 1)$
$\Rightarrow {(2 x^{2} - x - 6)}^{2} - {(2 x^{2} - 9 x + 10)}^{2}$
$∵ a^{2} - b^{2} = (a + b) (a - b)$
$\Rightarrow (2 x^{2} - x - 6 + 2 x^{2} - 9 x + 10) (2 x^{2} - x - 6 - 2 x^{2} + 9 x - 10)$
$\Rightarrow (4 x^{2} - 10 x + 4) (8 x - 16)$
$\Rightarrow 2 (2 x^{2} - 5 x + 2) 8 (x - 2)$
$\Rightarrow 2 (2 x^{2} - 4 x - x + 2) 8 (x - 2)$
$\Rightarrow 2 (2 x (x - 2) - 1 (x - 2)) 8 (x - 2)$
$\Rightarrow 16 {(x - 2)}^{2} (2 x - 1)$

Suggest Corrections

Similar questions

Find the factors of $x^{2} + 22 x + 112$ by splitting the middle term.

2 x^{2} + 9 x + 17 = x^{2} + 16 x + 5

x^{2} 2^{2 x} + 9 (x + 2) 2^{x} + 8 x^{2} ⩽ (x + 2) 2^{2 x} + 9 x^{2} 2^{x} + 8 x + 16

f (x) = ∣ ∣ ∣ ∣ \begin{matrix} x^{2} - 4 x + 6 & 2 x^{2} + 4 x + 10 & 3 x^{2} - 2 x + 16 x - 2 & 2 x + 2 & 3 x - 1 1 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣

x - 4

∣ ∣ ∣ ∣ \begin{matrix} x - 2 & 2 x - 3 & 3 x - 4 x - 4 & 2 x - 9 & 3 x - 16 x - 8 & 2 x - 27 & 3 x - 64 \end{matrix} ∣ ∣ ∣ ∣

Join BYJU'S Learning Program