Polynomials — CBSE Class 10 Maths Textbook Solutions

The zeroes of a polynomial $p(x)$p(x) are exactly the $x$x-coordinates of the points where the graph of $y=p(x)$y=p(x) meets the $x$x-axis — so the number of zeroes equals the number of points at which the curve touches or crosses the $x$x-axis. Reading each of the six graphs on this basis:

(i) The graph never touches the $x$x-axis at all, so $p(x)$p(x) has no zeroes.

(ii) The graph cuts the $x$x-axis at exactly one point, so $p(x)$p(x) has $1$1 zero.

(iii) The graph cuts the $x$x-axis at three points, so $p(x)$p(x) has $3$3 zeroes.

(iv) The graph cuts the $x$x-axis at two points, so $p(x)$p(x) has $2$2 zeroes.

(v) The graph cuts the $x$x-axis at four points, so $p(x)$p(x) has $4$4 zeroes.

(vi) The graph cuts the $x$x-axis at three points, so $p(x)$p(x) has $3$3 zeroes.

This is consistent with the rule that a polynomial of degree $n$n has at most $n$n zeroes — graph (v), which crosses the $x$x-axis four times, corresponds to a polynomial of degree $4$4 or higher.

For each quadratic $ax^2+bx+c$ax^2+bx+c, we factorise by splitting the middle term to get the zeroes, then check that $\alpha+\beta=-\dfrac{b}{a}$+=-b/a and $\alpha\beta=\dfrac{c}{a}$=c/a.

(i) $x^2-2x-8$x^2-2x-8
We need two numbers with product $-8$-8 and sum $-2$-2: these are $-4$-4 and $2$2.
$x^2-2x-8 = x^2-4x+2x-8 = x(x-4)+2(x-4) = (x-4)(x+2)$x^2-2x-8 = x^2-4x+2x-8 = x(x-4)+2(x-4) = (x-4)(x+2)
Zeroes: $x=4$x=4 and $x=-2$x=-2.
Sum $=4+(-2)=2=-\dfrac{(-2)}{1}=-\dfrac{b}{a}$=4+(-2)=2=-(-2)/1=-b/a. Product $=4\times(-2)=-8=\dfrac{-8}{1}=\dfrac{c}{a}$=4×(-2)=-8=-8/1=c/a. Both check out.

(ii) $4s^2-4s+1$4s^2-4s+1
We need two numbers with product $4\times1=4$4×1=4 and sum $-4$-4: these are $-2$-2 and $-2$-2.
$4s^2-4s+1 = 4s^2-2s-2s+1 = 2s(2s-1)-1(2s-1) = (2s-1)(2s-1)$4s^2-4s+1 = 4s^2-2s-2s+1 = 2s(2s-1)-1(2s-1) = (2s-1)(2s-1)
Zeroes: $s=\dfrac12$s=12 and $s=\dfrac12$s=12 (equal zeroes).
Sum $=\dfrac12+\dfrac12=1=-\dfrac{(-4)}{4}=-\dfrac{b}{a}$=12+12=1=-(-4)/4=-b/a. Product $=\dfrac12\times\dfrac12=\dfrac14=\dfrac{1}{4}=\dfrac{c}{a}$=12×12=14=1/4=c/a. Both check out.

(iii) $6x^2-3-7x$6x^2-3-7x
Rewrite in standard form: $6x^2-7x-3$6x^2-7x-3. We need two numbers with product $6\times(-3)=-18$6×(-3)=-18 and sum $-7$-7: these are $-9$-9 and $2$2.
$6x^2-7x-3 = 6x^2-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)$6x^2-7x-3 = 6x^2-9x+2x-3 = 3x(2x-3)+1(2x-3) = (3x+1)(2x-3)
Zeroes: $x=-\dfrac13$x=-13 and $x=\dfrac32$x=32.
Sum $=-\dfrac13+\dfrac32=\dfrac{-2+9}{6}=\dfrac{7}{6}=-\dfrac{(-7)}{6}=-\dfrac{b}{a}$=-13+32=-2+9/6=7/6=-(-7)/6=-b/a. Product $=-\dfrac13\times\dfrac32=-\dfrac12=\dfrac{-3}{6}=\dfrac{c}{a}$=-13×32=-12=-3/6=c/a. Both check out.

(iv) $4u^2+8u$4u^2+8u
$4u^2+8u = 4u(u+2)$4u^2+8u = 4u(u+2)
Zeroes: $u=0$u=0 and $u=-2$u=-2.
Sum $=0+(-2)=-2=-\dfrac{8}{4}=-\dfrac{b}{a}$=0+(-2)=-2=-8/4=-b/a. Product $=0\times(-2)=0=\dfrac{0}{4}=\dfrac{c}{a}$=0×(-2)=0=0/4=c/a. Both check out.

(v) $t^2-15$t^2-15
Using $a^2-b^2=(a-b)(a+b)$a^2-b^2=(a-b)(a+b):
$t^2-15 = \left(t-\sqrt{15}\right)\left(t+\sqrt{15}\right)$t^2-15 = (t-√15)(t+√15)
Zeroes: $t=\sqrt{15}$t=√15 and $t=-\sqrt{15}$t=-√15.
Sum $=\sqrt{15}+\left(-\sqrt{15}\right)=0=-\dfrac{0}{1}=-\dfrac{b}{a}$=√15+(-√15)=0=-0/1=-b/a. Product $=\sqrt{15}\times\left(-\sqrt{15}\right)=-15=\dfrac{-15}{1}=\dfrac{c}{a}$=√15×(-√15)=-15=-15/1=c/a. Both check out.

(vi) $3x^2-x-4$3x^2-x-4
We need two numbers with product $3\times(-4)=-12$3×(-4)=-12 and sum $-1$-1: these are $-4$-4 and $3$3.
$3x^2-x-4 = 3x^2-4x+3x-4 = x(3x-4)+1(3x-4) = (3x-4)(x+1)$3x^2-x-4 = 3x^2-4x+3x-4 = x(3x-4)+1(3x-4) = (3x-4)(x+1)
Zeroes: $x=\dfrac43$x=43 and $x=-1$x=-1.
Sum $=\dfrac43+(-1)=\dfrac13=-\dfrac{(-1)}{3}=-\dfrac{b}{a}$=43+(-1)=13=-(-1)/3=-b/a. Product $=\dfrac43\times(-1)=-\dfrac43=\dfrac{-4}{3}=\dfrac{c}{a}$=43×(-1)=-43=-4/3=c/a. Both check out.

If a quadratic polynomial has zeroes with sum $S$S and product $P$P, it can be written as
$k\left(x^2-Sx+P\right),\quad k\neq0,$k(x^2-Sx+P), k≠0,
since this expands to $x^2-(\alpha+\beta)x+\alpha\beta$x^2-(+)x+ for any zeroes $\alpha,\beta$,. Taking $k=1$k=1 (or a value that clears fractions/surds) gives one valid answer in each case.

(i) $S=\dfrac14,\ P=-1$S=14, P=-1:
$x^2-\dfrac14x-1$x^2-14x-1
Multiplying by $4$4 (taking $k=4$k=4) to clear the fraction: $\boxed{4x^2-x-4}$4x^2-x-4.

(ii) $S=\sqrt2,\ P=\dfrac13$S=2, P=13:
$x^2-\sqrt2x+\dfrac13$x^2-2x+13
Multiplying by $3$3: $\boxed{3x^2-3\sqrt2x+1}$3x^2-32x+1.

(iii) $S=0,\ P=\sqrt5$S=0, P=5:
$\boxed{x^2+\sqrt5}$x^2+5

(iv) $S=1,\ P=1$S=1, P=1:
$\boxed{x^2-x+1}$x^2-x+1

(v) $S=-\dfrac14,\ P=\dfrac14$S=-14, P=14:
$x^2+\dfrac14x+\dfrac14$x^2+14x+14
Multiplying by $4$4: $\boxed{4x^2+x+1}$4x^2+x+1.

(vi) $S=4,\ P=1$S=4, P=1:
$\boxed{x^2-4x+1}$x^2-4x+1

In every case, any nonzero multiple $k$k of the boxed polynomial is also a correct answer, since the question asks for a quadratic polynomial, not a unique one.