Question

If x^3-ax+b and x^2 - cx+ d are divisible by x-m, then prove that m = d - b / c - a​

Answer 1

Your Question :-

• $\bf{If \:{x}^{ \red{3}}-ax+b \:and\: x² - cx+ d\: are\:divisible\: by\: x-m, \:then \:prove \:that\: m = \frac{ d - b} {c - a}}$

But there is an mistake in your question the equation should be.

Question :-

• $\bf{If \:{x}^{ \green{2}}-ax+b \:and\: x² - cx+ d\: are\: divisible\: by\: x-m, \:then \:prove \:that\: m = \frac{ d - b} {c - a}}$

$\underline{\displaystyle{\green{\bold{Solution}:-}}}$

$\: \: \: \: \: \: \: \: \: \: \mathrm{x {}^{2} - ax + b } - - ( {eq}^{n} \: 1) \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \mathrm{x {}^{2} - cx + d } - - ( {eq}^{n} \: 2) \: \: \:$

$\red{\bigstar}$]Both Equation are divisible by x - m [Given in qu.]

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{x - m = 0} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \rightarrow \boxed{ \bold{ x = m}} \: \: \: \: \: \:$

Now, Putting x = m in both 1 & 2 equation.

$\begin{aligned}\cancel{ {m}^{2}} - am+b& =0\\ \cancel{{m}^{2}} - cm +d & =0 \\ \: \: \: \: \: \underline{\: \: \: \: \:- \: \: \: \: \:+ \: \: \: \: \: -\: \: \: \: \: \:} \end{aligned}$

(-a +c ) m + b - d = 0

( c - a ) m + b-d = 0

( c-a ) m = d - b

$\implies \boxed{\underline{\bf{\green{m = \frac{d - b}{c - a}}}}}$

Hence Proved .