Question

X-1 is a factor of x^3+7x^2+ax-3 than fine the value of a

Answer 1

Step-by-step explanation:

Answer:-

Given that:-

$\sf{x-1}$ is a factor of $\sf{x^3+7x^2+ax-3}$

To find:-

The value of a.

Concept:-

Placement of x in the polynomial, equating to zero.

Let's Do!

$\sf{x-1=0}$

$\sf{x=1}$

Now, placement:-

$\sf{x^3+7x^2+ax-3=0}$

$\sf{(1)^3+7(1)^2+a(1)-3=0}$

$\sf{1+7+a-3=0}$

$\sf{8+a-3=0}$

$\sf{5+a=0}$

$\sf{a= -5}$

Remember that:-

• The equation is equal to zero.
• This is because x-1 is the factor if the equation, hence, zero.

Answer 2

Given :

$\sf x-1 \: is \: a \: factor \: of \: x³+7x²+ax-3$

$\sf \implies x - 1 = 0$

$\sf \implies x = 1$

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To find :

$\sf value \: of \: a$

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Solution :

$\sf x³+7x²+ax-3=0$

$\sf Now, \: put \: value \: of \: x = 1$

$\sf (1)³+7(1)²+a(1) - 3=0$

$\sf 1+7× 1+a - 3=0$

$\sf 1+7+a-3=0$

$\sf 8+a-3=0$

$\sf 5+a=0$

$\boxed{\sf a= -5}$

$\sf \color{fuchsia} Therefore, \: value \: of \: a = -5$