Solve the polynomial using factorisation method: $(x^{2} - 20 x + 19)^{4}$

((x - 1) (x - 19))^{2}

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∣ ∣ ((x + 1) (x - 19))^{2} ∣ ∣

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| ((x - 1) (x - 19)) |

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∣ ∣ (x - 1)^{2} (x - 19)^{2} ∣ ∣

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Solution

The correct option is C $∣ ∣ (x - 1)^{2} (x - 19)^{2} ∣ ∣$
$\sqrt{(x^{2} - 20 x + 19)^{4}}$
Factor of $x^{2} - 20 x + 19 = (x - 1) (x - 19)$
$\sqrt{(x^{2} - 20 x + 19)^{4}} = \sqrt{(((x - 1) (x - 19))^{2})^{2}}$
$=$ $∣ ∣ (x - 1)^{2} (x - 19)^{2} ∣ ∣$

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Similar questions

Factorise $x^{2} - 3 x + 2$ by using the Factor Theorem.

Factorise $x^{2} + 3 x + 2$ by using the Factor Theorem.

Q. Prove the following

(1) {sin}^{- 1} (\frac{2 x}{1 + x^{2}}) = 2 {tan}^{- 1} x, | x | \leq 1

(2) {cos}^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = 2 {tan}^{- 1} x, x \geq 0

(3) {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) = 2 {tan}^{- 1} x, - 1 < x < 1

Q. Solve:

\int_{1}^{2} \frac{x}{(x + 1) (x + 2)} d x

Solve the following equations for x:
(i) tan⁻¹2x + tan⁻¹3x = nπ + $\frac{3 π}{4}$

(ii) tan⁻¹(x + 1) + tan⁻¹(x − 1) = tan⁻¹ $\frac{8}{31}$

(iii) $\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}$

(iv) sin⁻¹x + sin⁻¹2x = $\frac{π}{3}$

(v) $3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}$

(vi) $\cos^{- 1} x + \sin^{- 1} \frac{x}{2} = \frac{π}{6}$

(vii) tan⁻¹(x −1) + tan⁻¹x tan⁻¹(x + 1) = tan⁻¹3x

(viii) tan (cos⁻¹x) = sin $(\cot^{- 1} \frac{1}{2})$

(ix) tan⁻¹ $(\frac{1 - x}{1 + x}) - \frac{1}{2}$ tan⁻¹x = 0, where x > 0

(x) cot⁻¹x − cot⁻¹(x + 2) = $\frac{π}{12}$ , x > 0

(xi) $\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0$

(xii) tan⁻¹(x + 2) + tan⁻¹(x − 2) = tan⁻¹ $(\frac{8}{79})$ , x > 0

(xiii) $\tan^{- 1} \frac{x}{2} + \tan^{- 1} \frac{x}{3} = \frac{π}{4}, 0 < x < \sqrt{6}$

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Solve the polynomial using factorisation method: (x2−20x+19)4

The correct option is C ∣∣(x−1)2(x−19)2∣∣√(x2−20x+19)4Factor of x2−20x+19=(x−1)(x−19)√(x2−20x+19)4=√(((x−1)(x−19))2)2= ∣∣(x−1)2(x−19)2∣∣

Solve the polynomial using factorisation method: $(x^{2} - 20 x + 19)^{4}$

The correct option is C $∣ ∣ (x - 1)^{2} (x - 19)^{2} ∣ ∣$
$\sqrt{(x^{2} - 20 x + 19)^{4}}$
Factor of $x^{2} - 20 x + 19 = (x - 1) (x - 19)$
$\sqrt{(x^{2} - 20 x + 19)^{4}} = \sqrt{(((x - 1) (x - 19))^{2})^{2}}$
$=$ $∣ ∣ (x - 1)^{2} (x - 19)^{2} ∣ ∣$