# Polynomials

## Short Questions

If a factor of 2x^{3} - 6x^{2} - 5m - 2 is x - 2, find the value of m.

Solution,

Let P(x) = 2x^{3} - 6x^{2} - 5m - 2

And d(x) = x - 2

Comparing d(x) with x - a, we get, a = 2.

Here, d(x) is factor of P(x).

Then, by the converse of factor theorem,

P(a) = 0

or, P(2) = 0

or, 2(2)^{3} - 6(2)^{2} - 5m - 2 = 0

or, 16 - 24 - 5m - 2 = 0

or, -10 = 5m

∴ m = -2

If (x + 2) is a factor of the polynomial 2x^{3} + 3x^{2} - mx + 4, find the value of m.

Solution,

Let P(x) = 2x^{3} + 3x^{2} - mx + 4

And d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2.

Here, d(x) is factor of P(x).

Then, by the converse of factor theorem,

P(a) = 0
or, P(-2) = 0

or, 2(-2)^{3} + 3(-2)^{2} - m(-2) + 4 = 0

or, -16 + 12 + 2m + 4 = 0

or, 2m = 0

∴ m = 0

If 2x^{3} - 4x^{2} + kx + 10 is divided by (x + 2), then remainder os 4. Find the value
of k using
remainder theorem.

Polynomial p(x) = 2x

^{3}- 4

^{2}+ kx + 10

Divisor d(x) = x + 2

Remainder (R) = 4

Comparing d(x) with x - a, we get, a = -2

By remainder theorem,R = p(a)

or, R = p(-2)

or, 4 = 2(-2)

^{3}- 4(-2)

^{2}+ k(-2) + 10

or, 4 = -16 - 16 - 2k + 10

or, 20 + 16 = -2k + 10

or, 26 = -2k

∴ k = -13

If x - 3 is a factor of 2x^{3} - x^{2} + 10x - k then find the value of k.

Solution,

Let P(x) = 2x^{3} - x^{2} + 10x - k

And d(x) = x - 3

Comparing d(x) with x - a, we get, a = 3.

Here, d(x) is factor of P(x).

Then, by the converse of factor theorem,

P(a) = 0

or, P(3) = 0

or, 2(3)^{3} - 3^{2} + 10(3) - k = 0

or, 54 - 9 + 30 - k = 0

or, 75 - k = 0

∴ k = 75

If x - 5 is a factor of x^{3} + px^{2} +4x + 5, find the value of p.

Solution,

Let P(x) = x^{3} + px^{2} + 4x + 5

And d(x) = x - 5

Comparing d(x) with x - a, we get, a = 5.

Here, d(x) is a factor of P(x).

Then, by the converse of factor theorem,

P(a) = 0

or, P(5) = 0

or, (5)^{3} + p(5)^{2} + 4(5) + 5 = 0

or, 125 + 25p + 20 + 5 = 0

or, 150 = -25p

∴ p = -6

Using Remainder theorem, find the remainder when 8x^{3} - 4x^{2} + 2x - 5 is divided by
2x - 1.

p(x) = 8x

^{3}- 4x

^{2}+ 2x - 5

d(x) = 2x - 1

Comparing d(x) with ax - b, we get, a = 2 and b = 1.

By remainder theorem,

Remainder = p$\left(\frac{b}{a}\right)$

= p$\left(\frac{1}{2}\right)$

= 8$\left(\frac{1}{2}\right)^3$ - 4$\left(\frac{1}{2}\right)^2$ + 2$\left(\frac{1}{2}\right)$ - 5

= 1 - 1 + 1 - 5

= -4

If x - 5 is a factor of 2x^{3} - 7x + (p - 12), find the value of p.

If p(x) = 2x

^{3}- 7x + p - 12

And d(x) = x - 5

Comparing d(x) with x - a, we get a = 5

Here, d(x) is a factor of P(x)

Then, by the converse of factor theorem,

p(a) = 0

or, p(5) = 0

or, 2(5)^{3} - 7(5)p + p - 12 = 0

or, 250 - 35p + p - 12 = 0

or, 238 - 34p = 0

∴ p = 7

If x^{3} - 21x - 20 = (x + 1).q(x), find q(x) by using synthetic division method.

x

^{3}- 21x - 20 = (x + 1).q(x)

Comparing with, p(x) = d(x).q(x), we get,

p(x) = x^{3} - 21x - 20

d(x) = x + 1

Comparing d(x) with x - a, we get, a = -1

Then, using synthetic division method,
∴ q(x) = x^{2} - x - 20

If x^{3} - 19x - 30 = (x + 2).q(x), find q(x) by using synthetic division method.

Given,

x^{3} - 19x - 30 = (x + 2).q(x)

Comparing with p(x) = d(x).q(x), we get,

p(x) = x^{3} - 19x - 30

d(x) = (x + 2)

Comparing d(x) with x - a, we get, a = -2

Then, using synthetic division method,
∴ q(x) = x^{2} - 2x - 15

State factor theorem. Use factor theorem to determine whether x + 3 is a factor of the polynomial
x^{3} - 8x
+ 3.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,

p(x) = x^{3} - 8x + 3

d(x) = x + 3

Comparing d(x) with x - a, we get, a = -3

Now, to determine factor,p(a) = p(-3)

= (-3)

^{3}- 8(-3) + 3

= -27 + 24 + 3

= 0

Here, p(-3) = 0, that is remainder = 0. So, by factor theorem d(x) is a factor of p(x).

State the remainder theorem. When a polynomial x^{3} + 5x^{2} - 6x - 16 is divided by (x
+ 2), find
the remainder by applying remainder theorem.

Remainder theorem: If P(x) is a polynomial of degree n and (x - a) is a divisor of P(x) then P(a) is a remainder, where the degree of quotient will be (n - 1).

Given,

p(x) = x^{3} + 5x^{2} - 6x - 16

d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2

Now, By remainder theorem,

Remainder = p(a)

= p(-2)

= (-2)^{3} + 5(-2)^{2} - 6(-2) - 16

= -8 + 20 + 12 - 16

= 8

∴ Remainder = 8

If the polynomial x^{3} + 6x^{2} + kx + 10 is divided by (x + 2), the remainder is 4,
find the value
of k using the remainder theorem.

p(x) = x

^{3}+ 6x

^{2}+ kx + 10

d(x) = x + 2

Remainder (R) = 4

Comparing d(x) with x - a, we get, a = -2

Using remainder theorem,

R = p(a)
or, 4 = p(-2)

or, 4 = (-2)^{3} + 6(2)^{2} - 2k + 10

or, 12 = 24 - 2k + 10

or, -22 = -2k

∴ k = 11

If (x + 2) is a factor of x^{3} - 19x - p, find the value of p.

p(x) = x

^{3}- 19x - p

d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2

Here, d(x) is a factor of p(x).

Then, by the converse of factor theorem,

p(a) = 0

or, p(-2) = 0

or, (-2)^{3} - 19(-2) - p = 0

or, p = -8 + 38

∴ p = 30

If (x - m) is a factor of x^{3} - mx^{2} - 4x + m + 12, calculate the value of m.

p(x) = x

^{3}- mx

^{2}- 4x + m + 12

d(x) = x - m

Comparing d(x) with x - a, we get, a = m.

Here, d(x) is factor of p(x).

Then, by the converse of factor theorem,

p(a) = 0

or, p(m) = 0

or, m^{3} - m^{3} - 4m + m + 12

or, -3m = -12

∴ m = 4

Define factor theorem. If x - 2 is a factor of x^{3} - kx^{2} + x - 6 = 0, find the
value of 'k'.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,p(x) = x

^{3}- kx

^{2}+ x - 6 = 0

d(x) = x - 2

Comparing d(x) with x - a, we get, a = 2

Here, d(x) is a factor of p(x).

Then, by the converse of factor theorem,

p(a) = 0

or, p(2) = 0

or, (2)^{3} - k(2)^{2} + 2 - 6 = 0

or, 8 - 4k - 4 = 0

or, 4 = 4k

∴ k = 1

State factor theorem. Use factor theorem to determine whether x + 4 is a factor of the polynomial
x^{3} -
15x + 4.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,p(x) = x

^{3}- 15x + 4

d(x) = x + 4

Comparing d(x) with x - a, we get, a = -4

Now,

p(a) = p(-4)

= (-4)^{3} -
15(-4) + 4

= -64 + 60 + 4

= 0

Here, p(a) = remainder = 0. So, using factor theorem d(x) is a factor of p(x).

State remainder theorem. Use it to find the remainder when the polynomial x^{6} - 1 is divided
by x + 1.

Remainder theorem: If P(x) is a polynomial of degree n and (x - a) is a divisor of P(x) then P(a) is a remainder, where the degree of quotient will be (n - 1).

Solution,p(x) = x

^{6}- 1

d(x) = x + 1

Comparing d(x) with x - a, we get, a = -1

Now, using remainder theorem,

Remainder = p(a)

= p(-1)

= (-1)^{6} - 1

= (-1)^{2}.(-1)^{2}.(-1)^{2} - 1

= 1.1.1 - 1

∴ Remainder = 0

The polynomial f(x) = x^{3} - (P - 2)x^{2} - Px + 28 leaves a remainder 10 when divided
by x + 3.
Find the value of P, using the remainder theorem.

f(x) = x

^{3}- (P - 2)x

^{2}- Px + 28

Remainder(R) = 10

d(x) = x + 3

Comparing d(x) with x - a, we get, a = -3,

Using remainder theorem,R = f(a)

or, 10 = f(-3)

or, 10 = (-3)

^{3}- (P - 2)(-3)

^{2}- Px + 28

or, 10 = -27 -(P - 2)9 + 3P + 28

or, 37 = -9P + 18 + 3P + 28

or, 9 = 18 - 6P

or, -9 = -6P

∴ P = $\frac{3}{2}$