Question

if 1 and -3 are the zeros of the polynomial,p(x) = x^3-ax^2-13x+b, find the value of a and b

Answer 1

Given: p(x) = x^3 - ax^2 - 13x + b, 1 and -3 are the zeroes of p(x).

To find: The value of a and b.

Answer:

Let us first find the value of p(x) when x = 1 and equate it to zero.

p(1) = (1)^3 - a*(1)^2 - 13*(1) + b

p(1) = 1 - a - 13 + b

p(1) = -12 - a + b

Equating it to zero,

-12 - a + b = 0

-a + b = 12

b = 12 + a (Take this as equation 1.)

Now, let's find the value of (x) when x = -3 and equate it to zero.

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

p(-3) = -27 - 9a + 39 + b

p(-3) = 12 - 9a + b

Equating it to zero,

12 - 9a + b = 0

-9a + b = -12

Substituting the value of b in equation 1 to the above equation,

-9a + 12 + a = -12

-8a = -12 - 12

-8a = -24

a = -24/-8 = 3

Now that we know the value of a, let's substitute it in equation 1.

b = 12 + 3

b = 15

Therefore, a = 3 and b = 15

Answer 2

$\huge\orange{\boxed{\green{\mathbb{\underbrace{\overbrace{\fcolorbox{blue}{navy}{\underline{\color{ivory}{꧁Explanation :-꧂ }}}}}}}}}$

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

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

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

The value of a and b.

Finding the value of polynomial_{p(x)}

p(x)

( x = 1 ) :

• $\begin{gathered}\quad{\longmapsto \: \: \tt{p(1) = (1)^3 - a × (1)^2 - 13 × (1) + b}}.\\\end{gathered}$
• $\begin{gathered}\quad{\longmapsto \: \: \tt{p(1) = 1 - a - 13 + b}}.\\\end{gathered}$
• $\begin{gathered}\quad{\longmapsto \: \: \tt{p(1) = - 12 - a + b}}.\\\end{gathered}$

Now,

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

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

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

[ Let it be Equation 1 ]

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

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

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

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

Now,

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

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

[ Let it be Equation 2 ]

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

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

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

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

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

Then, substituting it in ( 1 ) :

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

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

∴ Values of :

a ⇝ 3

b ⇝ 15

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