Exercise 2.2 Practice
Zeroes of a Polynomial
Q1: Value of Polynomial
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Find the value of the polynomial $5x - 4x^2 + 3$ at:
(i) $x = 0$
(ii) $x = -1$
(iii) $x = 2$
(i) $x = 0$
(ii) $x = -1$
(iii) $x = 2$
Let $p(x) = 5x - 4x^2 + 3$.
(i) At $x = 0$:
$p(0) = 5(0) - 4(0)^2 + 3 = 3$.
$p(0) = 5(0) - 4(0)^2 + 3 = 3$.
(ii) At $x = -1$:
$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = -6$.
$p(-1) = 5(-1) - 4(-1)^2 + 3 = -5 - 4(1) + 3 = -5 - 4 + 3 = -6$.
(iii) At $x = 2$:
$p(2) = 5(2) - 4(2)^2 + 3 = 10 - 4(4) + 3 = 10 - 16 + 3 = -3$.
$p(2) = 5(2) - 4(2)^2 + 3 = 10 - 4(4) + 3 = 10 - 16 + 3 = -3$.
(i) 3, (ii) -6, (iii) -3
Q2: Find p(0), p(1), p(2)
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Find $p(0), p(1)$ and $p(2)$ for each of the following polynomials:
(i) $p(y) = y^2 - y + 1$
(ii) $p(t) = 2 + t + 2t^2 - t^3$
(iii) $p(x) = x^3$
(iv) $p(x) = (x-1)(x+1)$
(i) $p(y) = y^2 - y + 1$
(ii) $p(t) = 2 + t + 2t^2 - t^3$
(iii) $p(x) = x^3$
(iv) $p(x) = (x-1)(x+1)$
(i) $p(y) = y^2 - y + 1$:
$p(0) = 0 - 0 + 1 = 1$
$p(1) = 1 - 1 + 1 = 1$
$p(2) = 4 - 2 + 1 = 3$
$p(0) = 0 - 0 + 1 = 1$
$p(1) = 1 - 1 + 1 = 1$
$p(2) = 4 - 2 + 1 = 3$
(ii) $p(t) = 2 + t + 2t^2 - t^3$:
$p(0) = 2 + 0 + 0 - 0 = 2$
$p(1) = 2 + 1 + 2 - 1 = 4$
$p(2) = 2 + 2 + 8 - 8 = 4$
$p(0) = 2 + 0 + 0 - 0 = 2$
$p(1) = 2 + 1 + 2 - 1 = 4$
$p(2) = 2 + 2 + 8 - 8 = 4$
(iii) $p(x) = x^3$:
$p(0) = 0$, $p(1) = 1$, $p(2) = 8$.
$p(0) = 0$, $p(1) = 1$, $p(2) = 8$.
(iv) $p(x) = (x-1)(x+1) = x^2 - 1$:
$p(0) = -1$
$p(1) = 0$
$p(2) = 3$
$p(0) = -1$
$p(1) = 0$
$p(2) = 3$
Q3: Verify Zeroes
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Verify whether the following are zeroes of the polynomial, indicated against them.
(i) $p(x) = 3x + 1, x = -1/3$
(ii) $p(x) = 5x - \pi, x = 4/5$
(iii) $p(x) = x^2 - 1, x = 1, -1$
(iv) $p(x) = (x+1)(x-2), x = -1, 2$
(v) $p(x) = x^2, x = 0$
(vi) $p(x) = lx + m, x = -m/l$
(vii) $p(x) = 3x^2 - 1, x = -1/\sqrt{3}, 2/\sqrt{3}$
(viii) $p(x) = 2x + 1, x = 1/2$
(i) $p(x) = 3x + 1, x = -1/3$
(ii) $p(x) = 5x - \pi, x = 4/5$
(iii) $p(x) = x^2 - 1, x = 1, -1$
(iv) $p(x) = (x+1)(x-2), x = -1, 2$
(v) $p(x) = x^2, x = 0$
(vi) $p(x) = lx + m, x = -m/l$
(vii) $p(x) = 3x^2 - 1, x = -1/\sqrt{3}, 2/\sqrt{3}$
(viii) $p(x) = 2x + 1, x = 1/2$
(i) $p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0$. Yes.
(ii) $p(4/5) = 5(4/5) - \pi = 4 - \pi \neq 0$. No.
(iii) $p(1) = 1-1=0$; $p(-1) = 1-1=0$. Yes, both.
(iv) $p(-1) = 0(-3)=0$; $p(2) = (3)(0)=0$. Yes, both.
(v) $p(0) = 0^2 = 0$. Yes.
(vi) $p(-m/l) = l(-m/l) + m = -m + m = 0$. Yes.
(vii) $p(-1/\sqrt{3}) = 3(1/3) - 1 = 0$ (Yes).
$p(2/\sqrt{3}) = 3(4/3) - 1 = 3 \neq 0$ (No).
$p(2/\sqrt{3}) = 3(4/3) - 1 = 3 \neq 0$ (No).
(viii) $p(1/2) = 2(1/2) + 1 = 2 \neq 0$. No.
Q4: Find Zero
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Find the zero of the polynomial in each of the following cases:
(i) $p(x) = x + 5$
(ii) $p(x) = x - 5$
(iii) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax, a \neq 0$
(vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.
(i) $p(x) = x + 5$
(ii) $p(x) = x - 5$
(iii) $p(x) = 2x + 5$
(iv) $p(x) = 3x - 2$
(v) $p(x) = 3x$
(vi) $p(x) = ax, a \neq 0$
(vii) $p(x) = cx + d, c \neq 0, c, d$ are real numbers.
To find zero, set $p(x) = 0$.
(i) $x + 5 = 0 \Rightarrow x = -5$.
(ii) $x - 5 = 0 \Rightarrow x = 5$.
(iii) $2x + 5 = 0 \Rightarrow x = -5/2$.
(iv) $3x - 2 = 0 \Rightarrow x = 2/3$.
(v) $3x = 0 \Rightarrow x = 0$.
(vi) $ax = 0 \Rightarrow x = 0$ (since $a \neq 0$).
(vii) $cx + d = 0 \Rightarrow x = -d/c$.
(i) -5, (ii) 5, (iii) -5/2, (iv) 2/3, (v) 0, (vi) 0, (vii) -d/c