Question

find the value of a and b px= x^4-5x³+4x²+ax+b and x-1 or x-2 are the factors of px
plz guys solve this it's urgent
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Answer 1

Answer:

a = 8,   b = - 8

Step-by-step explanation:

As x - 1 and x - 2 are root of the given equation it must satisfy all the condition which x satisfies. So, when x = 1, and x = 2, value of polynomial is 0.

  For x = 1,

⇒ x^4 - 5x^3 + 4x^2 + ax + b = 0

⇒ (1)⁴ - 5(1)³ + 4(1)² + a(1) + b = 0

⇒ 1 - 5 + 4 + a + b = 0

⇒ b = - a

    For x = 2,

⇒ (2)⁴ - 5(2)³ + 4(2)² + a(2) + b= 0

⇒ 16 - 40 + 16 + 2a + b = 0

⇒ 2a + b = 8

⇒ 2a - a = 0

⇒ a = 8  

       Hence, b = - a = - 8

Answer 2

Answer:-

$\red{\bigstar}$$\large\leadsto\boxed{\bf\purple{a = 8}}$

$\red{\bigstar}$$\large\leadsto\boxed{\bf\purple{b = -8}}$

• Given:-

$\sf{p(x) = x^4 - 5x^3 + 4x^2 + ax + b}$

• To Find:-

Value of a and b

• Solution:-

Given that, x-1 and x-2 are the factors or roots of the given polynomial.

Hence,

→ $\sf{x - 1}$

$\implies\bf\red{x = 1}$

also,

→ $\sf{x - 2}$

$\implies\bf\red{x = 2}$

As these are the roots of the polynomial therefore they should satisfy the conditions as that of x.

$\pink{\bigstar}$ Taking $\bf\red{x = 1}$

Substituting the value in the polynomial:-

→ $\sf{(1)^4 - 5(1)^{3} + 4 (1)^{2} + a(1) + b = 0}$

→ $\sf{1 - 5 + 4 + a + b = 0}$

→ $\sf{5 -5 + a + b = 0}$

→ $\sf{a + b = 0}$

→ $\bf\green{a = - b}$

$\pink{\bigstar}$ Taking $\bf\red{x = 2}$

Substituting the value in the polynomial:-

→ $\sf{(2)^4 - 5(2)^{3} + 4 (2)^{2} + a(2) + b = 0}$

→ $\sf{16 - 5 \times 8 + 4 \times 4 + 2a + b = 0}$

→ $\sf{16 - 40 + 16 + 2a + b = 0}$

→ $\sf{32 - 40 + 2a + b = 0}$

→ $\sf{-8 + 2a + b = 0}$

→ $\sf{2a + b = 8}$

→ $\sf{2(-b) + b = 8}$$\: \: \: \: \:\longrightarrow\bf\red{[a = -b]}$

→ $\sf{-2b + b = 8}$

→ $\sf{-b = 8}$

→ $\bf\green{b = -8}$

Therefore,

$\large\boxed{\bf\pink{b = -8}}$

And

As, a = -b

Hence,

→ a = -(-8)

$\large\boxed{\bf\pink{a = 8}}$