Question

If 1 and -3 are zeroes of the polynomial p(x) = x^3 - ax^3 - 13x + b, find values of A and B ?
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Answer 1

Appropriate Question :-

If 1 and -3 are zeroes of the polynomial p( x ) = x³ - ax²-13x+b, find values of 'a' and 'b' ?

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Solution :-

Given Information :-

• p(x) = x³ - ax² - 13x+b,
• 1 and -3 are the zeroes of p(x)

To find :-

• The value of a and b.

Calculation :-

• Firstly, finding the value of p(x) when 'x = 1'

⇒ p(1) = (1)³ - a × (1)² - 13 × (1) + b

⇒ p(1) = 1 - a - 13 + b

⇒ p(1) = - 12 - a + b

Now, Equating it to zero, We get,

⇒ - 12 - a + b = 0

⇒ - a + b = 12

⇒ b = 12 + a -------------------- Equation1

• Finding the value of (x) when 'x = -3'

⇒ p(-3) = (-3)^3 - a × (-3)^2 - 13 × (-3) + b

⇒ p(-3) = -27 - 9a + 39 + b

⇒ p(-3) = 12 - 9a + b

Now, Equating it to zero, We get,

⇒ 12 - 9a + b = 0

⇒ -9a + b = -12 -------------------- Equation2

Substituting the value of Equation 1 in Equation 2, We get,

⇒ -9a + 12 + a = -12

⇒ -8a = -12 - 12

⇒ -8a = -24

⇒ $\bf a= \frac{ - 24}{ - 8} = 3$

∵ We know the value of a,

∴ We will substitute it in equation 1, We get,

⇒ b = 12 + 3

⇒ $\bf{b = 15}$

∴ Values of a and b :-

• a = 3
• b = 15

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Final Answer :-

• The value of a is 3
• The value of b is 15

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Answer 2

$\large{\underline{ \underline {\maltese \: \: {\textbf{\textsf{Explanation :-}}}}}}$

${ \underline {\maltese \: \: {\textrm{\textsf{Given Info. :-}}}}}$

$\quad\bullet \: \: \underline{ \boxed{\tt{p(x) = x {}^{3} - ax {}^{2} - 13x+b}}} \: \: \bigstar$

• 1 and -3 are the zeroes of the polynomial$_{p(x)}$.

${ \underline {\maltese \: \: {\textrm{\textsf{Exigency To find :-}}}}}$

• The value of a and b.

Finding the value of polynomial$_{p(x)}$, ( x = 1 ) :

$\quad{\longmapsto \: \: \tt{p(1) = (1)^3 - a × (1)^2 - 13 × (1) + b}}.$

$\quad{\longmapsto \: \: \tt{p(1) = 1 - a - 13 + b}}.$

$\quad{\longmapsto \: \: \tt{p(1) = - 12 - a + b}}.$

Now,

$\quad{\longmapsto \: \: \tt{- 12 - a + b = 0}}.$

$\quad{\longmapsto \: \: \tt{ a + b = 12 }}.$

$\quad{\longmapsto \: \: \tt{ b = 12 + a }}.$

[ Let it be Equation 1 ]

Finding the value of x (x = -3) :

$\quad{\longmapsto \: \: \tt{ p(-3) = (-3)^3 - a × (-3)^2 - 13 × (-3) + b }}.$

$\quad{\longmapsto \: \: \tt{ p(-3) = -27 - 9a + 39 + b }}.$

$\quad{\longmapsto \: \: \tt{ p(-3) = 12 - 9a + b }}.$

Now,

$\quad{\longmapsto \: \: \tt{ 12 - 9a + b = 0 }}.$

$\quad{\longmapsto \: \: \tt{ -9a + b = -12 }}.$

[ Let it be Equation 2 ]

Substituting the known value of ( 1 ) in ( 2 ) :

$\quad{\longmapsto \: \: \tt{ -9a + 12 + a = -12 }}.$

$\quad{\longmapsto \: \: \tt{ -8a = -12 - 12 }}.$

$\quad{\longmapsto \: \: \tt{ -8a = -24 }}.$

$\quad{\longmapsto \: \: \sf{ a= \frac{ - 24}{ - 8} = 3 }}.$

Then, substituting it in ( 1 ) :

$\quad{\longmapsto \: \: \tt{ b = 12 + 3 }}.$

$\quad{\longmapsto \: \: \sf{ b = 15 }}.$

∴ Values of :

• a ⇝ 3
• b ⇝ 15

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