Question

if (x-2)(x-4) are the factors of x³-ax²+14x+b then find values of a and b
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Answer 1

Answer:

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Answer 2

(x-2) and (x-4) are the factor of x³ - ax² + 14x + b

now find the value of a and b

to find first equation we know that

divider is (x-2)and dividend is x³ - ax² + 14x + b

so put (x-2) equal to zero

x-2 = 0

x=2

now put the value of x in the dividend

P(2) = 2³ - a×2² + 14×2 + b

= 8 - 4a + 28 + b

so

4a-b = 8 + 28

4a-b = 36 this is equation 1

now find equation 2

the divider is now (x-4)

so put (x-4) equal to zero

(x-4) = 0

x=4

now put the value of x in dividend

P(4) = 4³ - a×4² + 14×4 + b

= 64 - 16a + 56 + b

so

16a-b = 64 + 56

16a-b = 120 this is equation 2

now subtract equation 1 from equation 2

16a-b = 120

- 4a + b = -36 ( sign change because we subtract the equation)

so the value of a from the solving of equations is

12a = 84

a = 7

now we found the value of a

now we have to find the value of B

put the value of a in equation 1

4a - b = 36

4×7-b = 36

28 - b = 36

-b = 8

so

b = - 8

Verify the answer

4×7 - (-8) = 36 equation 1 verified

16×7 - (-8) = 120 equation 2 verified

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