Question

1.   If x = 2/3 and x = - 3 are the zeroes of the polynomial
  P(x) = A x2 + 7x + B then find the values of    A and B​

Answer 1

Given :

• $\longrightarrow \sf x = (-3) \: and \dfrac{2}{3} \: are \: zeroes \: of \: the \: polynomial \: p(x)$

• p(x) = Ax² + 7x + B

Aim :

• To find the values of A and B respectively.

Answer :

Factor Theorem :

According to the Factor Theorem, For any polynomial p(x), if p(a) = 0, then (x-a) is a factor of the polynomial.

Solution :

By using factor theorem, let us find the values respectively.

p(x) = Ax² + 7x + B

p(-3) :

⇒ A(-3)² + 7(-3) + B = 0

⇒ 9A - 21 + B = 0

$\sf p\bigg(\dfrac{2}{3}\bigg) :$

$\implies \sf A\bigg(\dfrac{2}{3} \bigg)^2 + 7\bigg(\dfrac{2}{3} \bigg) + B = 0$

$\implies \sf \dfrac{4A}{9}+ \dfrac{14}{3} + B = 0$

LCM of 9,3,1 = 9

$\implies \sf \dfrac{4A}{9} + \dfrac{14\times3}{3\times3} + \dfrac{B\times9}{1\times9} = 0$

$\implies \sf \cfrac{4A}{9} + \dfrac{42}{9} + \dfrac{B}{9} = 0$

Adding,

$\implies \sf \dfrac{4A + 42 + B}{9} = 0$

$\implies \sf 4A + 42 + B = 0\times 9 \rightarrow 0$

Since both the equations result in zero, we can equate both the equations.

⇒ 9A - 21 + B = 4A + 42 + B

Transposing all the constants to one side and variables to the other,

⇒ 9A + B - 4A - B = 28 + 42

⇒ 5A = 70

Transposing 5 to the other side,

$\implies \sf A = \dfrac{70}{5} \rightarrow 14$

Hence the value of A = 14

By substituting the value of A as 14 in one of the equations, let us find the value of B.

⇒ 4A + 42 + B = 0

⇒ 4(14) + 42 + B = 0

⇒ 56 + 42 + B = 0

⇒ 98 + B = 0

⇒ B = (-98)

Therefore, the Value of A and B respectively are 14 and (-98)

More to Know :

Remainder Theorem :

The remainder theorem states that, for a polynomial f(x), and divisor g(x) = (x-b), p(b) yields the remainder.