Question

If (x - 1), (x + 2) and (x - 2) are factors of x3 + ax2 + bx + c, then the value of a, b and c.​

Answer 1

Step-by-step explanation:

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Answer 2

The values of a,b,and c are -1, -4, and 4 respectively.

Given:

• If (x - 1), (x + 2) and (x - 2) are factors of x³ + ax² + bx + c.

To find:

• Find the value of a, b and c.

Solution:

Concept/Theorem to be used:

• Apply factor theorem: It states that if x-a is a factor of polynomial p(x), then p(a)=0.
• Solve equations to find the values of a,b and c.

Step 1:

Find three equations by putting the values of x in polynomial.

Let

$p(x) = {x}^{3} + a {x}^{2} + bx + c$

Put x=1

$p(1) = {1}^{3} + a {(1)}^{2} + b(1)+ c$

or

$\bf a + b+ c = - 1...eq1$

put x=-2

$p( - 2) = {( - 2)}^{3} + a {( - 2)}^{2} + b( - 2)+ c$

or

$\bf 4a - 2b + c = 8...eq2$

put x=2

$p(2) = {2}^{3} + a {(2)}^{2} + b(2)+ c$

or

$\bf 4a + 2b + c = - 8...eq3$

Step 2:

Subtract equations 2 and 3.

$4a - 2b + c = 8 \\ 4a + 2b + c = - 8 \\ ( - ) \: ( - ) \: ( - ) \: \: ( + ) \\ - - - - - - - - - - \\ - 4b = 16$

or

$\bf \red{b = - 4}$

Step 3:

Put the value of b in eq1 and eq2

$a - 4 + c = - 1$

or

$\bf a + c = 3...eq4$

and

$4a - 2( - 4) + c = 8$

or

$\bf 4a + c = 0...eq5$

Subtract both eq4 and eq5

$a + c = 3 \\ 4a + c = 0 \\ ( - ) \: \: ( - ) \: \: ( - ) \\ - - - - - - - \\ - 3a = 3$

or

$\bf \green{a = - 1 }$

Step 4:

Put the value of a in eq4.

$- 1 + c = 3$

or

$\bf \pink{c = 4}$

Thus,

Values of a,b, and c are -1, -4, and 4 respectively.

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