1.

In each pair of polynomials given below, find the number to be subtracted from the first to get a polynomial for which the second is as factor. Find also the second factor of the polynomial got on subtracting the number.i. x2 – 3x + 5, x – 4 ii. x2 – 3x + 5, x + 4 iii. x2 + 5x – 7, x – 1 iv. x2 – 4x – 3, x – 1

Answer»

i. p(x) = x2 – 3x + 5

If x – 4 is a factor, p(4) = 0 

p(4) = (4)2 – 3 x 4 + 5 = 9

For x – 4 to become a factor of p (4) must be equal to zero.

For p (4) = 0 here we have to subtract 9 from p(x).

That is, add -9 to p(x) for (x – 4) become a factor.

\(\therefore\) p(x) = x2 – 3x + 5 – 9 = x2 – 3x – 4

x2 – 3x – 4 = (x – a) (x – b)

= x2 – (a + b) x + ab

a + b = 3

ab = –4

(a – b)2 = (a + b)2 – 4 ab

= (3)2 – 4x – 4 = 25

a – b = 5

a + b = 3

a = 4; b = –1

x2 – 3x – 4 = (x – 4)(x + 1)

Second factor is (x +1)

ii. p(x) = x2 – 3x + 5 

If x + 4 is a factor, p(–4) = 0

p(–4) = (–4)2 – 3x – 4 + 5 = 33

For x + 4 to become a factor of p (–4) must be equal to zero.

For p (–4) = 0 here we have to subtract 33 from p(x). 

That is, add –33 to p(x) for (x + 4) become a factor.

\(\therefore\) p(x) = x2 – 3x + 5 – 33 = x2 – 3x – 28

x2 – 3x – 28 = (x – a) (x – b)

= x2 – (a + b) x + ab

a + b = 3

ab = -28

(a – b)2 = (a + b)2 – 4ab

= (3)3 – 4x – 28 = 121

a – b= 11

a + b = 3

a = 7;

b = -4 x2 – 3x – 28 = (x – 7)(x + 4)

Second factor is (x – 7)

iii. p(x) = x2 + 5x – 7

If x – 1 is a factor, p(1) = 0

p(1) = (1)2 +5 x 1 – 7 = –1

For x – 1 to become a factor of p (1) must be equal to zero.

For p (1) = 0 here we have to subtract –1 from p(x).

That is, add 1 to p(x) for (x – 1) become a factor.

p(x) = x2 + 5x – 7 + 1 = x2 + 5x – 6

x2 + 5x – 6 = (x – a) (x – b)

a + b = 5

ab = –6

a = –6; b= 1

x2 + 5x – 6 = (x + 6)(x – 1)

Second factor is (x + 6)

iv. p(x) = x2 – 4x – 3

If x – 1 is a factor, p(1) = 0

p(1) = (1)2 – 4 x 1 – 3 = –6

For x – 1 to become a factor of p (1) must be equal to zero.

For p (1) = 0 here we have to subtract -6 from p(x).

That is, add 6 to p(x) for (x – 1) become a factor.

\(\therefore\) p(x) = x2 – 4 x – 3 + 6 = x2 – 4x +3

x2 – 4x + 3 = (x – a) (x – b)

a + b = –4 ab = 3

a =1; b = 3

x2 – 4x +3 = (x – 1)(x – 3)

Second factor is (x – 3)


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