Question

if 2x²+x+5 and 2x²+4x-19 have a common factor(x-b),then show the value of b is equal to 8​

Answer 1

Solution :-

Let f(x) = 2x² + x + 5

and p(x) = 2x² + 4x - 19

Given

(x - b) is a factor of f(x) and g(x)

∴ By factor theorem

f(b) = 0 and p(b) = 0

i) Consider f(a) = 0

⇒ 2(b)² + b + 5 = 0

⇒ 2(b²) + b + 5 = 0

⇒ 2b² + b + 5 = 0 ----eq(1)

ii) Consider p(b) = 0

⇒ 2(b)² + 4(b) - 19 = 0

⇒ 2(b²) + 4b - 19 = 0

⇒ 2b² + 4b - 19 = 0 ----eq(2)

From eq(1) and eq(2)

⇒ 2b² + b + 5 = 2b² + 4b - 19

⇒ 2b² + b + 5 - 2b² - 4b + 19 = 0

⇒ - 3b + 24 = 0

⇒ 24 = 3b

⇒ 24/3 = b

⇒ 8 = b

⇒ b = 8

Hence shown

Answer 2

Given

$\sf 2x^2+x+5 \: and \:2x^2+4x-19\: have \\ \sf common \:factor (x-b)$

To show

$\boxed{\sf b=8}$

Explanation

$\sf as \:per \:the \:factor \:theorem,$

$\sf if\: (x-a) \:is \:a\: factor \:of \:p(x) \\ \\ \sf then \:p(a)=0$

$\sf let \:p(x)=2x^2+x+5 \\ \\ \sf g(x)=2x^2+4x-19$

Applying factor theorem,

Case 1:

$p(b) = 0$

substituting x=b

$2 {b}^{2} + b + 5 = 0..........(1)$

Case 2:

$g(b) = 0$

$2 {b}^{2} + 4b - 19 = 0...........(2)$

$\huge\undeline{\sf (1)=(2)}$

$2 {b}^{2} + b + 5 = 2 {b}^{2} + 4b - 19\\ \\ b-4b=-19-6$

$= > - 3b = - 24$

$= > \huge\boxed{ \sf \: b = 8 }$

Hence proved