+0  
 
0
414
3
avatar

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

Guest Aug 6, 2017

Best Answer 

 #3
avatar+93677 
+2

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

Melody  Aug 7, 2017
 #1
avatar+178 
+1

(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 \(273-39a=15\)

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

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

Jeffes02  Aug 7, 2017
 #2
avatar+20025 
+2

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

what is the value of a?

 

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

 

laugh

heureka  Aug 7, 2017
 #3
avatar+93677 
+2
Best Answer

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

Melody  Aug 7, 2017

16 Online Users

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.