Question

if the polynomial p(x)= x⁴-2x³+3x²-ax+8 is divided by x-2, it leaves a remainder 10. Find the value of a ?

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Answer 1

$\;\;\underline{\textbf{\textsf{First, let us know about polynomials }}}$

• Any expression which have more than two algebraic terms is know as

Polynomials.

Types of polynomials :-

• Monomial

• Binomial

• Trinomial

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$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• When p(x) = x⁴ - 2x³ + 3x² - ax + 8 divided by (x - 2), it leaves a remainder 10.

$\;\;\underline{\textbf{\textsf{ To find :-}}}$

• Value of " a "

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

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$\;\;\underline{\textbf{\textsf{ A.T.Q:-}}}$

$\\\\\longrightarrow \sf x - 2 = 0 \\\\\longrightarrow \boxed{\sf{\green{x = 2}}}$

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$\longrightarrow \sf x^4 - 2x^3 + 3x^2 - ax + 8 = 10 \\\\\\\qquad \quad \quad \underline{\sf{ Now \ substitute \ the \ Value \ of \ x \ in \ the \ given \ polynomial \ :}} \\\\\longrightarrow \sf (2)^4 - 2(2)^3 + 3(2)^2 - a(2) + 8 = 10 \\\\\longrightarrow \sf \cancel{16} - \cancel{16} + 12 - 2a + 8 = 10 \\\\\longrightarrow \sf 20 - 2a = 10 \\\\\longrightarrow \sf - 2a = 10 - 20 \\\\\longrightarrow \sf 2a = 10\\\\\\\longrightarrow \sf a = \cancel\dfrac{10}{2} \\\\\longrightarrow \boxed{\frak{\green{a = 5}}}$

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$\;\;\underline{\textbf{\textsf{ Hence -}}}$

${\therefore} \ \underline{\textsf{\textbf{ The \ Value \ of \ a \ is \ 5.}}}$

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Answer 2

$hello \: mate$

we have

p(x)=x^4–2x^3+3x^2-ax+b

By remainder theorem, when p(x) is divided by (x-1) and (x+1) , the remainders are equal to p(1) and p(-1) respectively.

By the given condition, we have

p(1)=5 and p(-1)=19

=> (1)^4–2(1)^3+3(1)^2-a(1)+b=5 and (-1)^4–2(-1)^3+3(-1)^2-a(-1)+b=19

=> 1–2+3-a+b=5 and 1-(-2)+3+a+b=19

=> -a+b=5–1+2–3 and 1+2+3+a+b=19

=> -a+b=3 and a+b=19–1–2–3

=> -a+b=3 and a+b=13

Adding these two equations,we get

-a+b+a+b=3+13

=> 2b=16

=> 2b/2=16/2

=> b=8

Putting b=8 in a+b=13 , we get

a+8=13

=> a=13–8

=> a=5

Therefore, a=5 and b=8 .