Question

(x-2) is a factor of x^3 + 2x^2 + ax - 2. Also verify if (x + 3) is also a factor

Answer 1

Answer :-

(x + 3) is a factor

Solution :-

Let f( x ) = x³ + 2x² + ax - 6

Given

(x - 2) is a factor of f( x )

Finding zero of x - 2

==> x - 2 = 0

==> x = 2

By Factor theorem

==> f( 2 ) = 0

==> ( 2 )³ + 2( 2 )² + a( 2 ) - 6 = 0

==> 8 + 2( 4 ) + 2a - 6 = 0

==> 8 + 8 + 2a - 6 = 0

==> 16 + 2a = 6

==> 2a = 6 - 16

==> 2a = - 10

==> a = - 5

So, the polynomial will be x³ + 2x² - 5x - 6

We need to check whether x + 3 is factor or not

When f( x ) is divided by (x + 3) if it leaves remainder 0 then (x + 3) will be the factor of f( x)

Finding zero of x + 3

==> x + 3 = 0

==> x = - 3

Finding remainder

By Remainder theorem

f( - 3 ) is the remainder

f( - 3 ) = ( - 3 )³ + 2( - 3 )²- 5( - 3 ) - 6

= - 27 + 2( 9 ) + 15 - 6

= - 27 + 18 + 15 - 6

= - 33 + 33

= 0

Since f( - 3 ) = 0 (x + 3) is the factor.

Answer 2

Given that,

(x - 2) is a factor of

${x}^{3} + 2 {x}^{2} + ax - 6$

if x -2 is a factor of given polynomial then

when we will put

x = 2 in the polynomial then it will become equal to zero.

so,

${x}^{3} + 2 {x}^{2} + ax - 6 \\ \\ {2}^{3} + 2 {(2)}^{2} + a(2)- 6 = 0 \\ \\ 8 + 8 + 2a - 6 = 0 \\ \\ 16 - 6 + 2a = 0 \\ \\ 2a \: = - 10 \\ \\ a = - 5$

so the polynomial will be

${x}^{3} + 2 {x}^{2} - 5x - 6$

now,

we have to verify if x = -3 is a factor of this polynomial so putting x = -3

${( - 3)}^{2} + 2 {( - 3)}^{2} - 5( - 3) - 6 \\ \\ - 27 + 18 + 15 - 6 \\ \\ - 33 + 33 \\ \\ = 0$

Since polynomial became equal to zero hence,

x +3 is a factor of polynomial.