Question

Let p(x) = 0 be a fifth degree polynomial equation with it's linear coefficents that has atleast 1 integral root .


If p(2) = 13 and p(10) = 5, Compute the value of x , that must satisfy p(x) = 0​

Answer 1

Let,

$\longrightarrow p(x)=(x-a)\,q(x)$

for $a\in\mathbb{Z},$ where $q(x)$ is a fourth degree polynomial having integer coefficients.

As per the question,

$\longrightarrow p(2)=13$

$\longrightarrow (2-a)\,q(2)=13$

Since $a\in\mathbb{Z},$ so are $2-a$ and $q(2)$ since $q(x)$ has integer coefficients. Thus we have two possibilities.

• $1\times13=13$

• $-1\times-13=13$

And also,

$\longrightarrow p(10)=5$

$\longrightarrow (10-a)\,q(10)=5$

Since $a\in\mathbb{Z},$ so are $10-a$ and $q(10)$ since $q(x)$ has integer coefficients. Thus again we have two possibilities.

• $1\times5=5$

• $-1\times-5=5$

We have to make cases by putting each possible value for $2-a,\ 10-a,\ q(2)$ and $q(10)$ from these possibilities to check where we get a valid value for $a.$

To avoid such cases we consider the following.

$\longrightarrow (2-a)-(10-a)=-8$

So we're sure $2-a$ and $10-a$ can't be $\pm1.$

Also,

$\longrightarrow (-13)-(-5)=-8$

Thus, without loss of generality, let,

$\longrightarrow2-a=-13\quad;\quad 10-a=-5$

Therefore, we get,

$\longrightarrow \underline{\underline{a=15}}$

I.e., $p(x)=0$ for $\bf{x=15.}$

Hence 15 is the answer.

Answer 2

Step-by-step explanation:

Answer:

$\huge {\purple {\fbox {\bigstar {\mathfb {\red {hello\:mate}}}}}}$

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$\huge {\purple {\fbox {\bigstar {\mathfb {\red {ur\: question}}}}}}$

$\huge {\mathfb{\purple {Q.}}}}$Let p(x) = 0 be a fifth degree polynomial equation with it's linear coefficents that has atleast 1 integral root .

If p(2) = 13 and p(10) = 5, Compute the value of x , that must satisfy p(x) = 0

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$\huge {\purple {\fbox {\bigstar {\mathfb {\red {ur\: ANSWER✓}}}}}}$

Let,

$\tt\implies p(x)= (x-a)×q(x)$

$\tt\implies (x-a)×q(x)$

$\red{\text{q(x)q(x)\: is\: a \:fourth\: degree\: polynomial\: having \:integer \:coefficients}}$ .

$\sf\green{\text{according\:to\:the\:question}}$

$\tt\implies (2-a)q(2)$

$\tt\implies =13(2-a)×q(2)=13$

$\green{\text{Since a\in\mathbb{Z},,\: so \:are \:(2-a)\:(2-a) \:and \:q(2)q(2).}}$

$\green{\text{since \:q(x)q(x) \:has \:integer\: coefficients}}$

$\purple{\text{Thus\: we\: have\: two\: possibilities}}$

$\tt\implies 1×13=13$

$\tt\implies (-1)×(-13)=13$

$\tt\red{\text{also,}}$

$\tt\implies p(10)=5$

$\tt\implies p(10)=5$

$\tt\implies (10-a)}q(10)$

$\tt\implies 5=>(10-a) q(10)=5$

$\pink{\text{Since \:a\in\mathbb{Z},,\: so \:are \:(10-a)(10-a)\: and\: q(10)q(10)}}$

$\pink{\text{since\: q(x)q(x) \:has\: integer\: coefficients}}$

$\pink{\text{. Thus\: again\: we \:have \:two\: possibilities.}}$

$\tt\implies 1×5=5$

$\tt\implies -1 × -5=5$

We have to make cases by putting each possible value for 2-a, 10-a, q(2)2-a, 10-a, q(2) and q(10)q(10) from these possibilities to check where we get a valid value for a.a.

$\pink{\text{to\: not\: get \:error\: consider \:the \:following :}}$

$\tt\implies (2-a)-(10-a)=-8$

$\sf\red{\text{so\:we\:now\:know}}$

$\tt\implies (2-a),(2-a) and (10-a)(10-a) can't\: be 1±1.$

$\tt\red{\text{and also}}$

$\tt\implies (-13)-(-5)=-8$

$\sf\red{\text{know\:let,}}$

$\tt\implies (2-a)=-13,\: (10-a)= -5$

$\tt\implies a=15$

$\blue{\text{I.e., p(x)=0 for \bf{x=15.}}}$

.

Hence

$\sf{\blue{\fbox{\purple{\bigstar{\mathbf{\red{therefore\:a=15}}}}}}}$

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$\huge\mathfb\red{hope\:it\:helps\:you}$