Question

If 2 and -3 are zeroes of polynomial
X°+(a+1)X+b find value of A and B

Answer 1

Question :-

If 2 and -3 are zeros of this polynomial then find the value of A and B .

$\sf{{x}^{2} + (a + 1)x + b}$

Answer :-

→ a = 0 ,b = -6

Solution :-

If the zeroes of any polynomial p(x) are giving then they will satisfy that polynomial and gives zero (0),

Let

$\sf{p(x) = {x}^{2} + (a + 1)x + b}$

If 2 is a zero of this therefore,

$\rightarrow \small \sf{p( 2)} = {( 2)}^{2} + (a + 1)( 2) + b = 0 \\ \\ \rightarrow \sf{0 = 4 + 2a + 2 + b} \\ \\ \rightarrow \sf{ 0 = 6 + 2a + b \: \: \: \: \: ........(1)}$

Now ,

If -3 is a zeros of the given polynomial,

then,

$\sf{p( - 3) = 0 = {( - 3)}^{2} + (a + 1)( - 3) + b} \\ \\ \rightarrow \sf{ 0 = 9 - 3a - 3 + b} \\ \\ \rightarrow \sf{ 0 = 6 - 3a + b} \: \: \: \: \: \: ......(2)$

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Now , eq.(1) - eq.(2)

→ 0 = 2a + 3 a

→ 0 = 5a

→ a = 0

Put a = 0 in eq.(2)

→ 0 = 6 - 0 +b

→ b = -6

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