Question

if both (x-2) and (2x-1) are factors of px^2+5x+r show that p/r=1

Answer 1

HERE IS YOUR ANSWER

$p(x) = p {x}^{2} + 5x + r$

$x - 2 = 0 \\ x = 2 \\ \\ 2x - 1 = 0 \\ x = \frac{1}{2}$

Put x = 2 in P(x)

$p {x}^{2} + 5x + r = 0 \\ p {(2)}^{2} + 5(2) + r = 0 \\ 4p + r + 10 = 0$

Put x = 1/2 in P(x)

$p {x}^{2} + 5x + r = 0 \\ p { \frac{1}{2} }^{2} + 5 \times \frac{1}{2} + r = 0 \\ \frac{1}{4} p + r + \frac{5}{2} = 0$

Subtracting the equations

$4p + r + 10 - ( \frac{1}{4} p + r + \frac{5}{2} ) = 0 \\ 4p + r + 10 - \frac{1}{4} p - r - \frac{5}{2} = 0 \\ \frac{15}{4} p + \frac{15}{2} = 0 \\ \frac{15}{4} p = - \frac{15}{2} \\ p = - 2$

$4p + r + 10 = 0 \\ 4( - 2) + r + 10 = 0 \\ - 8 + r + 10 = 0 \\ r + 2 = 0 \\ r = - 2$

$\frac{r}{p} = 1 \\ \\ \frac{ - 2}{ - 2} = 1 \\ 1 = 1$

Hence, proved.

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Answer 2

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✺Required Answer:

♦️ GiveN:

• $\large{\rm{(x-2) }}$and $\large{\rm{(2x-1)}}$ are the factors of polynomial f(x) = $\large{\rm{px^{2}+ 5x+r}}$

♦️ To ProvE:

• p/r = 1

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✺Explanation of Q.

The above question is based upon the application of factors theoram which is derived from the remainder theorems only, So let's see it's statement,

♠️Factor Theoram:-

Let f(x) be a polynomial of degree $\large{\rm{\geqslant}}$ 1 and a be any real constant such that f(a) = 0, then (x-a) is a factor of f(x). We can say conversely, if (x-a) is a factor of f(x), then f(a)=0. Actually in factor theorem, the remainder is 0, That's why, it is remainder theoram only with remainder = 0

♠️ Roots of any polynomial:-

A real number $\large{\rm{\alpha}}$ is a root of a polynomial of f(x)=0 if f($\large{\rm{\alpha}}$) i.e. $\large{\rm{\alpha}}$ is a zero of the polynomial f(x).

Now, let's solve this question by using these concepts : )

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✺Solution:-

Let f(x) = $\large{\rm{px^{2}+5x+r}}$

Given (x - 2) is a factor of polynomial f(x)

So, by using factor's theoram,

$\large{ \rm{ \dashrightarrow \: f(2) = 0 \: \red{(here \: \alpha = 2)}}} \\ \\ \large{ \rm{ \dashrightarrow{p( {2)}^{2}+ 5(2) + r = 0}}} \\ \\ \large{ \rm{ \dashrightarrow \: 4p + 10 + r = 0.........(1)}}$

Also given that (2x - 1) is a factor of f(x)

So, by using factor's theoram,

$\large{ \rm{ \dashrightarrow \: f( \frac{1}{2} ) = 0 \: \red{(here \: \alpha = \frac{1}{2})}}} \\ \\ \large{ \rm{ \dashrightarrow \: p( \frac{1}{2}) {}^{2} + 5( \frac{1}{2} ) + r = 0}} \\ \\ \large{ \rm{ \dashrightarrow \: \frac{p}{4} + \frac{5}{2} + r = 0}} \\ \\ \large{ \rm{ \dashrightarrow \: 4( \frac{p}{4} + \frac{5}{2} + r ) = 0 \times 4}} \\ \\ \large{ \rm{ \dashrightarrow \: p + 10 + 4r = 0.........(2)}}$

Subtracting eq.n(2) from eq.n(1),

$\large{ \rm{ \dashrightarrow \: 4p + 10 + r - (p + 10 + 4r) = 0}} \\ \\ \large{ \rm{ \dashrightarrow \: 4p + 10 + r - p - 10 - 4r = 0}} \\ \\ \large{ \rm{ \dashrightarrow \: 3p - 3r = 0}} \\ \\ \large{ \rm{ \dashrightarrow \: \cancel{3}p = \cancel{3}r}} \\ \\ \large{ \rm{ \dashrightarrow \: \boxed{ \rm{ \pink{\frac{p}{r} = 1}}}}} \\ \\ \large{ \therefore{ \rm{ \purple{ \underline{ \underline{Hence \: proved \: \dag}}}}}}$

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