Ch 2: Polynomials - Exercise 2.3 Practice

Q1: Check Factor (x+1)

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Determine which of the following polynomials has $(x + 1)$ a factor:
(i) $x^3 + x^2 + x + 1$
(ii) $x^4 + x^3 + x^2 + x + 1$
(iii) $x^4 + 3x^3 + 3x^2 + x + 1$
(iv) $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$

Let $p(x)$ be the polynomial. If $p(-1) = 0$, then $(x+1)$ is a factor.

(i) $x^3 + x^2 + x + 1$:
$p(-1) = (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0$.
Yes, $(x+1)$ is a factor.

(ii) $x^4 + x^3 + x^2 + x + 1$:
$p(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1 \neq 0$.
No.

(iii) $x^4 + 3x^3 + 3x^2 + x + 1$:
$p(-1) = 1 + 3(-1) + 3(1) - 1 + 1 = 1 - 3 + 3 - 1 + 1 = 1 \neq 0$.
No.

(iv) $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$:
$p(-1) = -1 - 1 - (2+\sqrt{2})(-1) + \sqrt{2} = -2 + 2 + \sqrt{2} + \sqrt{2} = 2\sqrt{2} \neq 0$.
No.

(i) Yes, (ii) No, (iii) No, (iv) No

Q2: Factor Theorem

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Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = 2x^3 + x^2 - 2x - 1, g(x) = x + 1$
(ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
(iii) $p(x) = x^3 - 4x^2 + x + 6, g(x) = x - 3$

(i) $g(x) = x+1 \Rightarrow x = -1$:
$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1 = -2 + 1 + 2 - 1 = 0$.
Yes, $g(x)$ is a factor.

(ii) $g(x) = x+2 \Rightarrow x = -2$:
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1 = -8 + 12 - 6 + 1 = -1 \neq 0$.
No, $g(x)$ is not a factor.

(iii) $g(x) = x-3 \Rightarrow x = 3$:
$p(3) = (3)^3 - 4(3)^2 + 3 + 6 = 27 - 36 + 3 + 6 = 36 - 36 = 0$.
Yes, $g(x)$ is a factor.

(i) Yes, (ii) No, (iii) Yes

Q3: Find Value of k

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Find the value of $k$, if $x - 1$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = x^2 + x + k$
(ii) $p(x) = 2x^2 + kx + \sqrt{2}$
(iii) $p(x) = kx^2 - \sqrt{2}x + 1$
(iv) $p(x) = kx^2 - 3x + k$

Since $x-1$ is a factor, $p(1) = 0$.

(i) $1^2 + 1 + k = 0 \Rightarrow 2 + k = 0 \Rightarrow k = -2$.

(ii) $2(1)^2 + k(1) + \sqrt{2} = 0 \Rightarrow 2 + k + \sqrt{2} = 0 \Rightarrow k = -(2 + \sqrt{2})$.

(iii) $k(1)^2 - \sqrt{2}(1) + 1 = 0 \Rightarrow k - \sqrt{2} + 1 = 0 \Rightarrow k = \sqrt{2} - 1$.

(iv) $k(1)^2 - 3(1) + k = 0 \Rightarrow 2k - 3 = 0 \Rightarrow k = 3/2$.

(i) -2, (ii) $-(2+\sqrt{2})$, (iii) $\sqrt{2}-1$, (iv) 3/2

Q4: Factorise (Quadratic)

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Factorise:
(i) $12x^2 - 7x + 1$
(ii) $2x^2 + 7x + 3$
(iii) $6x^2 + 5x - 6$
(iv) $3x^2 - x - 4$

(i) $12x^2 - 7x + 1$: Split -7x into -4x and -3x.
$12x^2 - 4x - 3x + 1 = 4x(3x - 1) - 1(3x - 1) = (3x - 1)(4x - 1)$.

(ii) $2x^2 + 7x + 3$: Split 7x into 6x and x.
$2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1)$.

(iii) $6x^2 + 5x - 6$: Split 5x into 9x and -4x.
$6x^2 + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2)$.

(iv) $3x^2 - x - 4$: Split -x into -4x and 3x.
$3x^2 - 4x + 3x - 4 = x(3x - 4) + 1(3x - 4) = (3x - 4)(x + 1)$.

Q5: Factorise (Cubic)

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Factorise:
(i) $x^3 - 2x^2 - x + 2$
(ii) $x^3 - 3x^2 - 9x - 5$
(iii) $x^3 + 13x^2 + 32x + 20$
(iv) $2y^3 + y^2 - 2y - 1$

(i) $x^3 - 2x^2 - x + 2$:
$= x^2(x - 2) - 1(x - 2) = (x^2 - 1)(x - 2) = (x + 1)(x - 1)(x - 2)$.

(ii) $x^3 - 3x^2 - 9x - 5$: Let $p(x)$. $p(-1) = -1 - 3 + 9 - 5 = 0$. So $(x+1)$ is a factor.
Divide $p(x)$ by $(x+1)$ to get $x^2 - 4x - 5$.
Factorise $x^2 - 4x - 5 = (x - 5)(x + 1)$.
Ans: $(x + 1)(x + 1)(x - 5) = (x + 1)^2(x - 5)$.

(iii) $x^3 + 13x^2 + 32x + 20$: $p(-1) = -1 + 13 - 32 + 20 = 0$. $(x+1)$ is a factor.
Divide to get $x^2 + 12x + 20$.
Factorise $x^2 + 12x + 20 = (x + 10)(x + 2)$.
Ans: $(x + 1)(x + 2)(x + 10)$.

(iv) $2y^3 + y^2 - 2y - 1$:
$= y^2(2y + 1) - 1(2y + 1) = (y^2 - 1)(2y + 1) = (y + 1)(y - 1)(2y + 1)$.