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Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x−1 and it leaves remainder 6 when divided by x+1; then :

Option 1)
p(2)=11

Option 2)
p(2)=19

Option 3)
p(−2)=19

Option 4)
p(−2)=11

Answers (2)

best_answer

As we have learned

Quadratic Expression -

$f\left ( x \right )= ax^{2}+bx+c$

- wherein

$a\neq 0$

$a,b,c\in R,\: \:$

Factor Theorem -

Any polynomial can be written in terms of product of its factors.

- wherein

If $P\left ( x \right )= 0$ has roots $\alpha _{1},\alpha _{2},\cdots \alpha _{n}$ and $P\left ( x \right )$ has leading coefficient ' $a$ ' then then $P\left ( x \right )= a\left ( x-\alpha _{1} \right )\left ( x-\alpha _{2} \right )\cdots \left ( x-\alpha _{n} \right )$

$p(x)= ax^2+bx+c$

$p(0)=C = 1$

$p(x)= ax^2 +bx +1$

$p(1)= a+b+1= 4$

$p(-1)= a-b+1= 6$

$2(a+1)= 10$

$\Rightarrow a= 4$

and

$\Rightarrow b= -1$

$p(x) = 4x^2-x+1$

$p(-z) = 19$

Option 1)

p(2)=11

Option 2)

p(2)=19

Option 3)

p(−2)=19

Option 4)

p(−2)=11

Posted by

Himanshu

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