Question

Find a and b such that x+1 and x+2 are factors of the polynomial x3 + ax2 -bx+10. Answer it plss

Answer 1

Hello friends!!

Here is your answer :

p(x) = x³ + ax² - bx + 10

g(x) = x + 1

Put g(x) = 0

x + 1 = 0

x = - 1

g(x) is a factor of p(x).

Therefore, p(-1) = 0

${( - 1)}^{3} + a{( - 1)}^{2} - b( - 1) + 10$


$- 1 + a + b + 10 = 0$

$a + b + 9 = 0$

$a = - 9 - b \: \: \: - - - - - - (1)$


_______________

f(x) = x + 2

Put f(x) = 0

x + 2 = 0

x = - 2

f(x) is a factor of p(x).

Therefore, p(-2) = 0

${( - 2)}^{3} + a {( - 2)}^{2} - b ( - 2) + 10 = 0$

$- 8 + 4a + 2b + 10 = 0$


$4a + 2b + 2 = 0$


Putting the value of a from eqⁿ (1)

$4( - 9 - b) + 2b + 2 = 0$


$- 36 - 4b + 2b + 2 = 0$

$- 34 - 2b = 0$

$- 2b = 34$



$b = \frac{ - 34}{2}$


$b = - 17$


Put the value of b in eqⁿ (1)

$a = - 9 - ( - 17)$


$a = - 9 + 17$


$a = 8$



Hope it helps you... ☺️☺️☺️☺️

# Be Brainly

Answer 2

Answer:

a = 2 and b = 13.

Step-by-step explanation:

Let p(x) = x3 + ax2 -bx+10

x+1 is a factor of p(x), so p(-1) = 0

-1 + a + b + 10 = 0 or a + b + 9 = 0                 ... (1)

x+2 is a factor of p(x), so p(-2) = 0

-8 + 4a + 2b + 10 = 0 or 2a + b + 1 = 0            ... (2)

Solving equations (1) and (2) to get the required values of a and b

we get

Let p(x) = x²+ax²-bx+ 10 and g(x) = x²-3x+2

Factorise g(x) = x²-3x+2:

x²-3x+2=x²-2x-x+2=x(x-2)-1(x-2)=(x-2)(x-1)

Therefore, g(x) = (x − 2)(x − 1)

It is given that p(x) is divisible by g(x), therefore, by factor theorem p(2) = 0 and p(1)=0. Let us first find p(2) and p(1) as follows:

p(1) = 13+ (a 12)-(bhp 10=1+ (a1)-b-10=a-b-11 p(2)=23+(a2)-(b2)-10=8+ (a4)-2b+10=4a-2b+18

Now equate p(2) = 0 and p(1) = 0 as shown below:

a-b-11-0

now equate p(2) = 0 and p(1) = 0 as shown below:

a-b+11=0

⇒a-b-11 .......(1)

4a-2b+18=0

=2(2a-b+9)=0

2a-b-9=0

2a-b-9

Now subtract equation 1 from equation 2:

(2a-a)+(b+b)=(-9-11)

a=2

Substitute a = 2 in equation 1:

2-b=-11

⇒-b=-11-2

⇒⇒b = −13 b=13

Hence, a = 2 and b = 13.

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