Question

8. Determine the value of 'a' when (x+3) is a factor of the polynomial p(t)=at^3 + t^2-22t-
21.
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Answer 1

GIVEN:-

• (x+3) is a factor of the polynomial p(t)=at³ + t^²- 22t - 21

TO FIND:-

• The value of "a".

CONCEPT USED:-

• ${\boxed{\rm{ Factor\:theorem}}}$.

• if (x - a) is a factor of p(x) then R = 0 and so on P(a) = 0.

Now,

$\implies\rm{ f(t) = (x + 3) = 0}$

$\implies\rm{ f(t) = x = -3}$.

Putting the value of "x" in (t)

$\implies\rm{p(t)=at^3 + t^2 - 22t - 21}$

$\implies\rm{ p(-3)a(-3)^3(-3)^2-22(-3) - 21}$

$\implies\rm{-27a + 9 + 66 - 21 = 0}$

$\implies\rm{-27a + 54 = 0}$

$\implies\rm{-27a = -54}$

$\implies\rm{a=\dfrac{54}{27}}$

$\implies\rm{a=3}$

Hence, The value of a is 3.

Answer 2

$\huge\bigstar\sf\pink{GIVEN}\bigstar$

(x+3) is a factor of the polynomial p(t)=at³ + t^²- 22t - 21

$\huge\bigstar\sf\pink{TO. FIND}\bigstar$

The value of "a".

if (x - a) is a factor of p(x) then R = 0 and so on P(a) = 0.

• factor theoram
• if (x - a) is a factor of p(x) then R = 0 and so on P(a) = 0.

$\huge\bigstar\sf\pink{NOW}\bigstar$

f(t)=x+3=0

f(t)=x=(-3)

PUTTING THE VALUE OF X IN T

p(t)=at^3+t^2-22t-21=p(-3)=a(-3)^3+(-3)^2-22(-3)-21

-27a+9+66-21=0

-27a+54=0

-27a=-54

a=54/27

a=3

SO the value of a = 3