Question

Find h and k so that p(x)= x - x² 14x + k has (x-1) and (x+3) as factors

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Answer 1

k=−3, zeroes of 2x

4

+x

3

−14x

2

+5x+6 are 1,−3,2,−

2

1

and zeroes of x

2

+2x−3 are 1,−3.

Since x

2

+2x+k is a factor of 2x

4

+x

3

−14x

2

+5x+6, so we dividing 2x

4

+x

3

−14x

2

+5x+6 by x

2

+2x+k should leave the remainder 0 as shown below:

Comparing the coefficient of x, we get

21+7k=0

⇒7k=−21

⇒k=−

7

21

=−3

The equation x

2

+2x+k becomes x

2

+2x−3 and the factors of x

2

+2x−3=0 are:

x

2

+3x−x−3=0

x(x+3)−1(x+3)=0

(x−1)(x+3)=0

x=1,−3

Now, the equation 2x

2

−3x−(8+2k) becomes 2x

2

−3x−2 and the factors of 2x

2

−3x−2=0 are:

2x

2

−4x+x−2=0

2x(x−2)+1(x−2)=0

(2x+1)(x−2)=0

x=−

2

1

,2

Hence, the zeros of the given two polynomials are 1,−3,−

2

1

and 2.