Question

if -1 and -2 are the zeros of the cubic polynomial x³ - 2 X² + a x + b, then find the values of a and b.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:a=-13}}}$

$\green{\tt{\therefore{Value\:of\:b=-10}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {x}^{3} - 2 {x}^{2} + ax + b = 0 \\ \\ \tt: \implies - 1 \: and \: - 2 \: are \: zeroes \: of \: cubic \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Value \: a = ? \\ \\ \tt: \implies Value \: b = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies p(x) = {x}^{3} - 2 {x}^{2} + ax + b = 0 \\ \\ \tt: \implies p( - 1) = { (- 1)}^{3} - 2( - 1)^{2} + a \times ( - 1) + b = 0 \\ \\ \tt: \implies - 1 - 2 - a + b = 0 \\ \\ \tt: \implies b - a = 3 - - - - - (1) \\ \\ \tt: \implies p( x) = {x}^{3} - 2 {x}^{2} + ax + b = 0 \\ \\ \tt: \implies p( - 2) = {( - 2)}^{3} - 2( - 2)^{2} + a \times ( -2) + b = 0 \\ \\ \tt: \implies - 8 - 8 - 2a + b = 0 \\ \\ \tt: \implies b - 2a = 16 - - - - - (2) \\ \\ \text{Subtracting \: (1) \: from \: (2)} \\ \tt: \implies b - 2a - b + a = 16 - 3 \\ \\ \tt: \implies - a = 13 \\ \\ \green{\tt: \implies a = - 13} \\ \\ \text{Putting \: value \: of \: a \: in \: (1)} \\ \tt: \implies b - ( - 13) = 3 \\ \\ \tt: \implies b + 13 = 3 \\ \\ \green{\tt: \implies b = - 10}$

Answer 2

Answer:

a = -13 b = 16

Step-by-step explanation:

x³-2x²+ax+b=p(x)

zero of p(x) is p(x) = 0

x = -1 (as given)

-1³-2*(-1²)+a*(-1)+b = 0

-1-2(1)+(-a)+b = 0

-1-2-a+b=0

-3-a+b = 0

-a+b = 3. equation 1

x = -2

p(x) = 0 when x is -2

put x = -2 in the polynomial

-2³-2*(-2²)+a*(-2)+b=0

-8-2*4-2a+b = 0

-8-8-2a+b = 0

-16-2a+b = 0

-2a+b = 16. equation 2

equating 1 and 2

-2a+b-(-a+b) = 16-3

-2a+b+a-b = 13

-2a+a+b-b = 13

-a = 13

a = -13

put a = -13 in eq.1

-(-13)+b = 3

13+b = 3

b = -10