Question

If (x+1) is a factor of
${x}^{2} - 3ax + 3a - 7$
then find the value of a.​

Answer 1

Step-by-step explanation:

above is your answer

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Answer 2

★ Concept ::

Here the concept of Factor Theorem has been used. We see that we are given a expression and there we need to find the value of a. So firstly using Factor Theorem, we can apply the value of x in the expression. And then equating this with 0 according to factor theorem, we can find the value of a.

Let's do it !!

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★ Solution :-

Given,

» Dividend = p(x) = x² - 3ax + 3a - 7

» Divisor = g(x) = x + 1

✒ Factor Theorem : By Factor Theorem, we know that if a polynomial g(x) is the factor of another polynomial p(x) then, when the value of x from g(x) is applied to p(x) then the result comes out to be 0.

From this, we get

→ g(x) = 0

→ x + 1 = 0

→ x = -1

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~ For the value of a ::

We have,

$\;\sf{\rightarrow\;\;p(x)\;=\;\bf{x^{2}\;-\;3ax\;+\;3a\;-\;7}}$

By applying the value of x here, we get

$\;\sf{\rightarrow\;\;p(-1)\;=\;\bf{(-1)^{2}\;-\;3(-1)a\;+\;3a\;-\;7}}$

$\;\sf{\rightarrow\;\;p(-1)\;=\;\bf{1\;-\;(-3)a\;+\;3a\;-\;7}}$

$\;\sf{\rightarrow\;\;p(-1)\;=\;\bf{1\;+\;3a\;+\;3a\;-\;7}}$

$\;\sf{\rightarrow\;\;\green{p(-1)\;=\;\bf{3a\;+\;3a\;-\;6}}}$

From Factor Theorem, we get when we apply the value of g(x) then, p(x) = 0.

So,

$\;\bf{\Longrightarrow\;\;3a\;+\;3a\;-\;6\;=\;0}$

$\;\bf{\Longrightarrow\;\;3a\;+\;3a\;=\;6}$

$\;\bf{\Longrightarrow\;\;6a\;=\;6}$

$\;\bf{\Longrightarrow\;\;a\;=\;\dfrac{6}{6}}$

$\;\bf{\red{\Longrightarrow\;\;a\;=\;1}}$

$\;\underline{\boxed{\tt{Hence,\;\;value\;\:of\;\:a\;=\;\bf{\purple{1}}}}}$

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★ Verification :-

For verifying we need to simply apply the value of a that we got into the equation we formed.

We have,

→ p(x) = x² - 3ax + 3a - 7

→ LHS = x² - 3ax + 3a - 7

→ RHS = 0 (from g(x))

• For LHS ::

✒ LHS = (-1)² - 3(1)(-1) + 3(1) - 7

✒ LHS = 1 + 3 + 3 - 7

✒ LHS = 7 - 7

✒ LHS = 0

We know that,

✒ RHS = 0

Clearly, LHS = RHS.

Hence, Verified.