Question

If 1 and -1 are the zeroes of the polynomial
p(x) = ax^3+ x² - 2x +b, find the values of 'a'and 'b'​

Answer 1

${EXPLANATION}$

Answer :-

• a= 2
• b = -1

Given :-

ax³ + x² - 2x + b have the zeros 1 , - 1

To find :-

Value of a, b

Solution:-

Since ,

1, - 1 are zeros If we substitute the values in place of x We get two equations From these two equations we can find the value of a, b

ax³ + x² - 2x + b

Substitute x = 1

a(1)³ +(1)² -2(1) + b =0

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

a + b - 1 =0------- eq 1

Now Substitute x = -1

ax³ + x² -2x + b =0

a(-1)³ +(-1)² -2(-1) + b =0

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

-a + 3 + b=0 ------ eq 2

Now subtracting both equations we get 1 - 2

a+ b - 1 -[-a + 3 + b] =0

a+ b -1 + a -3 - b = 0

2a -4 = 0

2a = 4

a= 2

Substitute value of a in eq 2

-a + 3 + b = 0

-2 + 3 + b = 0

b+1 = 0

b = -1

So, the value of a, b are 2 , - 1

Verification:-

Substitute value of a, b , x it should be equal to 0

Case - 1:-

• x = 1
• a = 2
• b = -1

ax³ + x² -2x + b = 0

2(1)³ +(1)² -2(1) -1 = 0

2 + 1 -2 -1 =0

3-3 =0

0= 0

Verified!

Case - 2 :-

• x = -1
• a = 2
• b = -1

ax³ + x² -2x + b = 0

2(-1)³ +(-1)² -2(-1) -1 =0

-2 +1+ 2 -1 =0

-3 +3 =0

0=0

Verified !