Question

Find the value of k if (x-1) is a factor of p(x)=2x
2+kx+5

Answer 1

Answer:

The value of ‘k’ is -7

Explanation:

Given equation,

$\rm \implies \: p(x) = {2x}^{2} + kx + 5$

Whose one of the factor is (x - 1)

By factor theorem,

$\rm \implies \: (x - 1) = 0$

$\rm \implies \: x - 1 = 0$

$\rm \implies \: x = 1$

Substituting p(1) as follows :-

$\rm \implies \: p(x) = {2x}^{2} + kx + 5$

$\rm \implies \: p(1) = {2(1)}^{2} + k(1) + 5$

Since, it's a factor p(x) = 0

Then,

$\rm \implies \: {2(1)}^{2} + k(1) + 5 = 0$

$\rm \implies \: {2(1)} + k(1) + 5 = 0$

$\rm \implies \: {2} + k + 5 = 0$

$\rm \implies \: k + 7 = 0$

$\rm \implies \: k = - 7$

${ \underline{ \rm \therefore \:The \: value \: of \: k\: is - 7}}$

__________________________

Quick check:

If ‘k' is -7

Then, the equation goes like this,

$\rm \implies \: p(x) = {2x}^{2} - 7x + 5$

If we substitute x as 1 we must get 0 as a result.

$\rm \implies \: p(x) = {2x}^{2} - 7x + 5$

$\rm \implies \: {2(1)}^{2} - 7(1) + 5 = 0$

$\rm \implies \: 2 - 7 + 5 = 0$

$\rm \implies \: 7 - 7 = 0$

$\rm \implies \: 0 = 0$

Hence, proved!

Answer 2

Answer:

Answer:

The value of ‘k’ is -7

Explanation:

Given equation,

\rm \implies \: p(x) = {2x}^{2} + kx + 5⟹p(x)=2x

2

+kx+5

Whose one of the factor is (x - 1)

By factor theorem,

\rm \implies \: (x - 1) = 0⟹(x−1)=0

\rm \implies \: x - 1 = 0⟹x−1=0

\rm \implies \: x = 1⟹x=1

Substituting p(1) as follows :-

\rm \implies \: p(x) = {2x}^{2} + kx + 5⟹p(x)=2x

2

+kx+5

\rm \implies \: p(1) = {2(1)}^{2} + k(1) + 5⟹p(1)=2(1)

2

+k(1)+5

Since, it's a factor p(x) = 0

Then,

\rm \implies \: {2(1)}^{2} + k(1) + 5 = 0⟹2(1)

2

+k(1)+5=0

\rm \implies \: {2(1)} + k(1) + 5 = 0⟹2(1)+k(1)+5=0

\rm \implies \: {2} + k + 5 = 0⟹2+k+5=0

\rm \implies \: k + 7 = 0⟹k+7=0

\rm \implies \: k = - 7⟹k=−7

{ \underline{ \rm \therefore \:The \: value \: of \: k\: is - 7}}

∴Thevalueofkis−7

__________________________

Quick check:

If ‘k' is -7

Then, the equation goes like this,

\rm \implies \: p(x) = {2x}^{2} - 7x + 5⟹p(x)=2x

2

−7x+5

If we substitute x as 1 we must get 0 as a result.

\rm \implies \: p(x) = {2x}^{2} - 7x + 5⟹p(x)=2x

2

−7x+5

\rm \implies \: {2(1)}^{2} - 7(1) + 5 = 0⟹2(1)

2

−7(1)+5=0

\rm \implies \: 2 - 7 + 5 = 0⟹2−7+5=0

\rm \implies \: 7 - 7 = 0⟹7−7=0

\rm \implies \: 0 = 0⟹0=0

Hence, proved!