P(x)= (x-a)(x-10)(x-20)+15. Given that p(7)=15.... what is the value of a?

P(x)= (x-a)(x-10)(x-20)+15. Given that p(7)=15.... what is the value of a?

Given

P(x)= (x-a)(x-10)(x-20)+15    and P(7)=15

then let

Q(x)= (x-a)(x-10)(x-20)     where we know    Q(7)=0

so 

x=7 is a root of Q so a=7

(Seems like no one want to answer the question after half a day then)

Your entry:P(x)=(x-a)(x-10)(x-20). Given that P(7)=15, what is the value of a?

Lets expand the polynomial P(x) first:

\(P(x)=\left(x-a\right)\left(x-10\right)\left(x-20\right)+15\)

\(=\left(x-a\right)\left(x^2-30x+200\right)\)

\(P(7)=15\)

Plug x=7 into P(x)

\(P(7)=(7-a)(7^2-30\cdot 7+200)=15\)

\(=(7-a)(49-210+200)\)

\(=39(7-a)\)

\(=273-39a\)

We know that \(273-39a=15\)

\(a=-(15-273)/39\)

\(a=-288/39=-96/13\)

P(x)= (x-a)(x-10)(x-20)+15. Given that p(7)=15....

what is the value of a?

\(\begin{array}{|rcll|} \hline P(x) &=& (x-a)(x-10)(x-20)+15 \quad & | \quad x = 7 \\\\ P(7) &=& (7-a)(7-10)(7-20)+15 \quad & | \quad P(7) = 15 \\ 15 &=& (7-a)(7-10)(7-20)+15 \quad & | \quad -15 \\ 0 &=& (7-a)(7-10)(7-20) \\\\ 0 &=& (7-a)(-3)(-13) \\ 0 &=& (7-a)\cdot 3\cdot 13 \\ 0 &=& (7-a)\cdot 39 \quad & | \quad :39 \\ 0 &=& (7-a)\cdot 1 \\ 0 &=& 7-a \\ a &=& 7 \\ \hline \end{array}\)