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

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

Aug 6, 2017

#3
+95017
+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

Aug 7, 2017

#1
+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 $$273-39a=15$$

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

$$a=-288/39=-96/13$$

.
Aug 7, 2017
#2
+20812
+2

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

what is the value of a?

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

Aug 7, 2017
#3
+95017
+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