Question

B. Use the Factor Theorem to determine whether or not the first polynomial is a
factor of the second.
1.
X - 1; x2 + 2x + 5
2. x + 1; x3 – X - 2
3. x - 4; 2x3 - 9x2 + 9x - 20
4.
a - 1; a3 - 2 a2 + a - 2
5. y + 3; 2y3 + y2 – 13y +6​

Answer 1

Step-by-step explanation:

Given:

1. x - 1; x² + 2x + 5

2. x + 1; x³ – x - 2

3. x - 4; 2x³ - 9x²+ 9x - 20

4. a - 1; a³ - 2 a² + a - 2

5. y + 3; 2y³ + y² – 13y +6

To find:Use the Factor Theorem to determine whether or not the first polynomial is a factor of the second.

Solution:

According to the statement of factor theorem if (x-a) is a factor of polynomial p(x),then p(a)=0.

Thus,apply the same for given cases.

1. x - 1; x² + 2x + 5

Put x=1 in the polynomial

$( {1)}^{2} + 2(1) + 5$

$1 + 2 + 5$

$7 \neq0$

Thus,

(x-1) is not a factor of x²+2x+5.

2. x + 1; x³ – x - 2

put x+1=0

x=-1

Find the value of polynomial at x= -1

$( { - 1)}^{3} - ( - 1) - 2$

$- 1 + 1 - 2$

$- 2 \neq0$

(x+1) is not a factor of x³ – x - 2.

3. x - 4; 2x³ - 9x²+ 9x - 20

put x=4

$2( {4)}^{3} - 9( {4) }^{2} + 9(4) - 20$

$2 \times 64 - 9 \times 16 + 36 - 20$

$128 - 144 + 36 - 20$

$164 - 164$

$= 0$

x - 4 is a factor of 2x³ - 9x²+ 9x - 20.

4. a - 1; a³ - 2 a² + a - 2

put a=1

$( {1)}^{3} - 2( {1)}^{2} + 1 - 2$

$1 - 2 + 1 - 2$

$2 - 4$

$- 2 \neq0$

a - 1 is not a factor of a³ - 2 a² + a - 2.

5. y + 3; 2y³ + y² – 13y +6

put y=-3

$2( { - 3)}^{3}+( { -3 )}^{2} – 13( - 3)+6$

$- 54 + 9 + 39 + 6$

$- 54 + 54$

$= 0$

y+3 is a factor of 2y³ + y² – 13y +6.

Hope it helps you.

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Answer 2

Step-by-step explanation:

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