Question

Find the value of a and b so that (x+1) & (x-2) are the factors of x^2+ax^2+2x+b

Answer 1

When x + 1 is a factor of x² + ax² + 2x + b = 0:

x = -1

p(x) = x² + ax² + 2x + b = 0

p(-1) => (-1)² + a(-1)² + 2(-1) + b = 0

=> 1 + a - 2 + b = 0

=> a + b = 1

=> a = 1 - b ___(1)

When x - 2 is a factor of x² + ax² + 2x + b = 0:

x = 2

p(2) => (2)² + a(2)² + 2(2) + b = 0

=> 4 + 4a + 4 + b = 0

=> 8 + 4a + b = 0

Put (1):

=> 8 + 4(1 - b) + b = 0

=> 8 + 4 - 4b + b = 0

=> 3b = 12

=> b = 4

a = 1 - b = 1 - 4

=> a = -3

Answer 2

Aɴꜱᴡᴇʀ

$\huge\sf{a= \:-3}$

$\huge\sf{b=4}$

_________________

Gɪᴠᴇɴ

$\sf{}(x + 1) \: and \: (x - 2) \: are \: the \: factors \: of \: {x}^{2} + a {x}^{2} + 2x + b$

_________________

ᴛᴏ ꜰɪɴᴅ

Value of a and b?????

_________________

Sᴛᴇᴘꜱ

$\large \sf{}if \: (x - 1) \: is \: a\: factor \: then \: equating \: this \: to \: 0 \\ \large\sf{} = {x +1 = 0} \\ \sf{} = x = \:- 1 \: \\ \large\sf{}substituting \:t his \: in \: {x}^{2} + a {x}^{2} + 2x + b \\ \large \sf{}p(1) = {(-1)}^{2} + a(-1) + 2(-1) + b \\ \large \sf{} a+b=1\: \: \\ \large \sf so \: a + b = 3\\ \sf a=1-b.....1$

$\large \sf {(x - 2) \: is \: a \: factor \: of \: {x}^{2} + a {x}^{2} + 2x + b } \\ \large \sf so \: x = 2 \\ \large \sf {2}^{2} + a {2}^{2} + 2(2) + b \\ \large \sf 4 + a {2}^{2} + 4+ b \\ \sf = 8+ 4a + b ..... ...........2$

So a=3-b

Sub this in 2 we get

8+4(1-b)+b=0

8+4-4b+b=0

12-3b=0

$\sf{-b\frac{ -12}{3}}$

b= 4

so sub this in 1

we get a=1-4

a= -3

_________________

$\huge{\mathfrak{\purple{hope\; it \;helps}}}$