Question

give full solution
question number 14
no irrelevant answers ​

Answer 1

$\huge \fbox \green{❤Answer:}$

Given:

${x}^{3} + {ax}^{2} - bx + 10$

$\textbf{is divisible by}$

${x}^{2} - 3x + 2$

$\text{To Find: Value of a and b}$

$\huge \fbox \pink{Solution:}$

${x}^{2} - 3x + 2 = (x - 2)(x - 1)$

$∴ {x}^{3} + {ax}^{2} - bx + 10$

$\text{is divisible by}$

$x - 2 \: and \: x - 1$

$\fbox{So,}$

$f(x) = {x}^{3} + {ax}^{2} - bx + 10$

$∴f(1) = 0 \: and \: f(2) = 0$

$⇒ {1}^{3} + a. {1}^{2} - b.1 + 10 = 0$

$⇒a≠b = - 11 - (1)$

$\fbox{Again,}$

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

$⇒2a - b = - 9 - (2)$

$⇒(2) - (1)→a = 2$

$∴b = 13$

Hence,

$:a = 2 ,b = 13$

$\\ \\ \\ \\ \sf \colorbox{gold} {\red(ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}$

Answer 2

Given:

{x}^{3} + {ax}^{2} - bx + 10x3+ax2−bx+10

\textbf{is divisible by}is divisible by

{x}^{2} - 3x + 2x2−3x+2

\text{To Find: Value of a and b}To Find: Value of a and b

\huge \fbox \pink{Solution:}Solution:

{x}^{2} - 3x + 2 = (x - 2)(x - 1)x2−3x+2=(x−2)(x−1)

∴ {x}^{3} + {ax}^{2} - bx + 10∴x3+ax2−bx+10

\text{is divisible by}is divisible by

x - 2 \: and \: x - 1x−2andx−1

\fbox{So,}So,

f(x) = {x}^{3} + {ax}^{2} - bx + 10f(x)=x3+ax2−bx+10

∴f(1) = 0 \: and \: f(2) = 0∴f(1)=0andf(2)=0

⇒ {1}^{3} + a. {1}^{2} - b.1 + 10 = 0⇒13+a.12−b.1+10=0

⇒a≠b = - 11 - (1)⇒a=b=−11−(1)

\fbox{Again,}Again,

{2}^{3} + a. {2}^{2} - b.2 + 10 = 023+a.22−b.2+10=0

⇒2a - b = - 9 - (2)⇒2a−b=−9−(2)

⇒(2) - (1)→a = 2⇒(2)−(1)→a=2

∴b = 13∴b=13

Hence,

:a = 2 ,b = 13:a=2,b=13

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