The two linear factors of $x^{2} - 15 x + 56$ are:

(x - 7)

and

(x + 8)

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(x - 7)

and

(x - 8)

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(x + 7)

and

(x - 8)

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(x + 7)

and

(x + 8)

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Solution

The correct option is C $(x - 7)$ and $(x - 8)$
$x^{2} - 15 x + 56 = 0$
$x^{2} - 8 x - 7 x + 56 = 0$
$x (x - 8) - 7 (x - 8) = 0$
$(x - 7) (x - 8) = 0$
$x = 7, 8$
$∴ (x - 7)$ and $(x - 8)$ are two linear factors.
So, Option $(b)$ is correct.

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Similar questions

8 x + 7 = 0

\Rightarrow 8 x = - 7 + 0

\Rightarrow 8 x = - 7

\Rightarrow x = 8 + 7

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8 x + 7 = 0

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and

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; find the value of:

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x^{2} + \frac{1}{x^{2}} = 7

and

x \neq 0

; find the value of :

7 x^{3} + 8 x - \frac{7}{x^{3}} - \frac{8}{x}

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The two linear factors of x2−15x+56 are:

The correct option is C (x−7) and (x−8)x2−15x+56=0x2−8x−7x+56=0x(x−8)−7(x−8)=0(x−7)(x−8)=0x=7,8∴(x−7) and (x−8) are two linear factors.So, Option (b) is correct.

The two linear factors of $x^{2} - 15 x + 56$ are:

The correct option is C $(x - 7)$ and $(x - 8)$
$x^{2} - 15 x + 56 = 0$
$x^{2} - 8 x - 7 x + 56 = 0$
$x (x - 8) - 7 (x - 8) = 0$
$(x - 7) (x - 8) = 0$
$x = 7, 8$
$∴ (x - 7)$ and $(x - 8)$ are two linear factors.
So, Option $(b)$ is correct.