# SOLUTION: Determine whether -3 and 2 are zeros of P(x) = 2x3 - 5x2 - 4x + 12. A. Neither B. Both C. Only -3 D. Only 2

Algebra ->  -> SOLUTION: Determine whether -3 and 2 are zeros of P(x) = 2x3 - 5x2 - 4x + 12. A. Neither B. Both C. Only -3 D. Only 2       Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Mathway solves algebra homework problems with step-by-step help!

Question 89990: Determine whether -3 and 2 are zeros of P(x) = 2x3 - 5x2 - 4x + 12.
A. Neither
B. Both
C. Only -3
D. Only 2

You can put this solution on YOUR website!
Lets find out if -3 is a zero

We can use synthetic division to find out if -3 is a zero

So let our test zero equal -3

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
 -3 | 2 -5 -4 12 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
 -3 | 2 -5 -4 12 | 2

Multiply -3 by 2 and place the product (which is -6) right underneath the second coefficient (which is -5)
 -3 | 2 -5 -4 12 | -6 2

Add -6 and -5 to get -11. Place the sum right underneath -6.
 -3 | 2 -5 -4 12 | -6 2 -11

Multiply -3 by -11 and place the product (which is 33) right underneath the third coefficient (which is -4)
 -3 | 2 -5 -4 12 | -6 33 2 -11

Add 33 and -4 to get 29. Place the sum right underneath 33.
 -3 | 2 -5 -4 12 | -6 33 2 -11 29

Multiply -3 by 29 and place the product (which is -87) right underneath the fourth coefficient (which is 12)
 -3 | 2 -5 -4 12 | -6 33 -87 2 -11 29

Add -87 and 12 to get -75. Place the sum right underneath -87.
 -3 | 2 -5 -4 12 | -6 33 -87 2 -11 29 -75

Since the last column adds to -75, we have a remainder of -75. This means -3 is not a zero

------------------------------------------------------

Lets find out if 2 is a zero

We can use synthetic division to find out if 2 is a zero

So let our test zero equal 2

Now set up the synthetic division table by placing the test zero in the upper left corner and placing the coefficients of the numerator to the right of the test zero.
 2 | 2 -5 -4 12 |

Start by bringing down the leading coefficient (it is the coefficient with the highest exponent which is 2)
 2 | 2 -5 -4 12 | 2

Multiply 2 by 2 and place the product (which is 4) right underneath the second coefficient (which is -5)
 2 | 2 -5 -4 12 | 4 2

Add 4 and -5 to get -1. Place the sum right underneath 4.
 2 | 2 -5 -4 12 | 4 2 -1

Multiply 2 by -1 and place the product (which is -2) right underneath the third coefficient (which is -4)
 2 | 2 -5 -4 12 | 4 -2 2 -1

Add -2 and -4 to get -6. Place the sum right underneath -2.
 2 | 2 -5 -4 12 | 4 -2 2 -1 -6

Multiply 2 by -6 and place the product (which is -12) right underneath the fourth coefficient (which is 12)
 2 | 2 -5 -4 12 | 4 -2 -12 2 -1 -6

Add -12 and 12 to get 0. Place the sum right underneath -12.
 2 | 2 -5 -4 12 | 4 -2 -12 2 -1 -6 0

Since the last column adds to zero, we have a remainder of zero. This means 2 is a zero