A little bit harder

Algebra Level 3

Let p ( x ) p(x) be a third degree polynomial such that p ( − 3 ) = p ( − 1 ) = p ( 2 ) = 0 p(-3) = p(-1) = p(2) = 0 and p ( 0 ) = 6 p(0) =6 . Find p ( 1 ) p(1) .

This problem is a part of this set .


The answer is 8.

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Sanchayapol Lewgasamsarn by Sanchayapol Lewgasamsarn , 20, Thailand

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1 solution

Ronak Agarwal
Aug 11, 2014

It is clear that − 1 , − 3 , 2 -1,-3,2 are the roots of the third degree polynomial.

So p ( x ) = a ( x + 1 ) ( x + 3 ) ( x − 2 ) p(x)=a(x+1)(x+3)(x-2)

Now given p ( 0 ) = 6 ⇒ 6 = − 6 a , a = − 1 p(0)=6 \quad \Rightarrow 6=-6a, a=-1

⇒ p ( x ) = ( x + 1 ) ( x + 3 ) ( 2 − x ) \Rightarrow p(x)=(x+1)(x+3)(2-x)

Put x = 1 x=1 to get p ( 1 ) = 8 p(1)=8

19 Helpful 0 Interesting 0 Brilliant 0 Confused

Did the same way.....BTW nice solution!!!

Harsh Shrivastava - 6 years, 10 months ago

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OH....it was so simple.......first I took f(x) = ax^3 + bx^2 + cx + d then found the value of a,b,c,d as -1,-2,5,6 using the determinant method and the found the value of f(1) as 8..... I never thought its solution would be so small and a bit tricky.....By the way, good solution..

Vighnesh Raut - 6 years, 6 months ago

very good solution!!

Mayank Holmes - 6 years, 10 months ago

Nice... Last years maths is all coming back to me now good solution

Zohaib Suleman - 5 years, 9 months ago

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