Question

If (x-1) is a factor of 2x3 + 8x2 +3x + k then find the value of k.​

Answer 1

Given:

• (x-1) is a factor of 2x³ + 8x² + 3x + k

Find:

• what will be the value of k?

Solution:

Here,

(x - 1) is a factor of eq. so,

Let, x - 1 = 0

x = 1

If we put x = 1 in eq. So, it gives us 0.

So,

$\rm \to2 {x}^{3} + 8 {x}^{2} + 3x + k = 0$

$\rm \to2 {(1)}^{3} + 8 {(1)}^{2} + 3(1) + k = 0$

$\rm \to2 (1)+ 8 (1) + 3 + k = 0$

$\rm \to2 + 8 + 3 + k = 0$

$\rm \to10 + 3+ k = 0$

$\rm \to13+ k = 0$

$\rm \to k = - 13$

Hence, the value of k will be -13.

VERIFICATION

Put value of k in 10 + 3 + k = 0 to verify our answer

$\rm \longrightarrow 10 + 3 + k = 0$

$\rm \longrightarrow 10 + 3 + ( - 13)= 0$

$\rm \longrightarrow 13 + ( - 13)= 0$

$\rm \longrightarrow \cancel{13} \cancel{- 13}= 0$

$\rm \longrightarrow 0= 0$

Hence, Verified

Answer 2

GIVEN :-

• (x - 1) is a factor of p(x) 2x³ + 8x² + 3x + k.

TO FIND :-

• The value of k.

SOLUTION :-

Let,

$\\ : \implies \displaystyle \sf x - 1 = 0$

$: \implies \displaystyle \sf x = 0 + 1$

$: \implies \underline{\boxed{\displaystyle \sf x = 1}}$

__________________

$\\ \\ \dashrightarrow \displaystyle \sf p(x) = 2x ^{3} + 8x ^{2} + 3x + k = 0$

$\displaystyle \sf \dashrightarrow \: p(1) = 2 \times (1) ^{3} + 8 \times (1) ^{2} + 3 \times 1 + k = 0$

$\dashrightarrow \displaystyle \sf \: 2 \times 1 + 8 \times 1 + 3 + k = 0$

$\dashrightarrow \displaystyle \sf2 + 8 + 3 + k = 0$

$\dashrightarrow \displaystyle \sf13 + k = 0$

$\dashrightarrow \underline{ \boxed{ \displaystyle \sf \: k = - 13}} \: \: \: \: \: \: \: \: \: \: \bigg \lgroup \displaystyle \sf \: value \: of \: k.\bigg \rgroup$

$\therefore \underline{ \displaystyle \sf The \ Value \ of \ k\ is \ -13}$