The factors of a quadratic polynomial $P (x) = x^{2} + 4 x - 5$ are:

(x + 5), (x - 1)

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(x + 5), (x + 1)

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(x + 1), (x - 5)

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(x - 1), (x - 5)

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Solution

The correct option is A $(x + 5), (x - 1)$
Given:
$P (x) = x^{2} + 4 x - 5$
$= x^{2} + 5 x - x - 5$
$= x (x + 5) - 1 (x + 5)$
$= (x + 5) (x - 1)$
$∴$ The factors of $P (x)$ are $(x + 5), (x - 1)$

Suggest Corrections

Similar questions

P (x) = x^{2} + 4 x - 5

- 5

\frac{- 1}{5}

\frac{x + 1}{x - 1} - \frac{x - 1}{x + 1} = \frac{5}{6}, x \neq 1, - 1

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

x = - 1

x = 5

p (x) = (x - c o s 36^{\circ}) (x - c o s 84^{\circ}) (x - c o s 156^{\circ})

x^{2}

\frac{1}{x - 1} - \frac{1}{x + 5} = \frac{6}{7}, x \neq 1, - 5

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