Chapter ISG, Problem 8.6.2.6P

a.

To determine

To find zeroes of polynomial

Answer to Problem 8.6.2.6P

  $x=2$ and $x=-1$ are the zeroes

Explanation of Solution

Given information:

  $y=-2x^2+2x+4$

Calculation:

To find zeroes,

y=0

Substituting value,

  $-2x^2+2x+4=0$

Now multiplying by -1,

  $2x^2-2x-4=0$

Dividing by 2,

  $x^2-x-2=0$

Now grouping,

  $x^2+x-2x-2=0$

Taking common,

  $x(x+1)-2(x+1)=0$

Taking common again,

  $(x+1)(x-2)=0$

Solving x,

  $x=2$ and $x=-1$

Hence, $x=2$ and $x=-1$ are the zeroes

b.

To determine

To find zeroes of polynomial

Answer to Problem 8.6.2.6P

Zeroes of equation are $x=\frac{5}{3}$ ,2.

Explanation of Solution

Giveninformation:

  $y=-3x^2+11x-10$

Calculation:

To find zeroes,

y=0

Substituting value,

  $-3x^2+11x-10=0$

Now multiplying by -1,

  $3x^2-11x+10=0$

Now,

  $3x^2-6x-5x+10=0$

Taking common,

  $3x(x-2)-5(x-2)=0$

Again taking common,

  $(3x-5)(x-2)=0$

Solving for x,

  $x-2=0$

  $3x-5=0$

Hence, zeroes of equation are $x=\frac{5}{3}$ ,2.