Polynomials — Important Questions

Answer: (c) 2 — The graph meets the $x$-axis at exactly two points, so $p(x)$ has $2$ real zeroes.

Answer: (a) $x^2+x-6$ — Sum $=2+(-3)=-1$ and product $=2\times(-3)=-6$. So $p(x)=x^2-(\text{sum})x+\text{product}=x^2-(-1)x+(-6)=x^2+x-6$.

Answer: (a) $\dfrac{3}{4}$ — Here $\alpha+\beta=6$ and $\alpha\beta=8$, so $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{6}{8}=\dfrac{3}{4}$.

Answer: (d) $-\dfrac{2}{3}$ — Sum $=-\dfrac{2}{k}$ and product $=\dfrac{3k}{k}=3$. Setting them equal: $-\dfrac{2}{k}=3\Rightarrow k=-\dfrac{2}{3}$.

Answer: (a) Both A and R are true and R correctly explains A. For $x^2+4$, $\ b^2-4ac=0-4(1)(4)=-16<0$, so its graph never meets the $x$-axis and it has no real zeroes.

$x^2-2x-8=(x-4)(x+2)$, so the zeroes are $x=4$ and $x=-2$.

Verification: Sum $=4+(-2)=2=-\dfrac{(-2)}{1}=-\dfrac{b}{a}$ ✓ and product $=4\times(-2)=-8=\dfrac{-8}{1}=\dfrac{c}{a}$ ✓.

The required polynomial is $k\left(x^2-(\text{sum})x+\text{product}\right)=k\left(x^2-\sqrt2\,x+\dfrac{1}{3}\right)$. Taking $k=3$ to clear the fraction gives $3x^2-3\sqrt2\,x+1$.

Splitting the middle term: $6x^2-7x-3=6x^2-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1)$.

Zeroes: $x=\dfrac{3}{2}$ and $x=-\dfrac{1}{3}$.

Verification: Sum $=\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{7}{6}=-\dfrac{(-7)}{6}=-\dfrac{b}{a}$ ✓; product $=\dfrac{3}{2}\times\left(-\dfrac{1}{3}\right)=-\dfrac{1}{2}=\dfrac{-3}{6}=\dfrac{c}{a}$ ✓.

Rewrite the polynomial: $x^2-p(x+1)-c=x^2-px-(p+c)$. Hence $\alpha+\beta=p$ and $\alpha\beta=-(p+c)$.

Now $(\alpha+1)(\beta+1)=\alpha\beta+(\alpha+\beta)+1=-(p+c)+p+1=1-c.$ Hence proved.

From the polynomial, $\alpha+\beta=5$; and it is given that $\alpha-\beta=1$.

Adding the two: $2\alpha=6\Rightarrow\alpha=3$, so $\beta=2$.

Then $k=\alpha\beta=3\times2=6$.

Since $\sqrt2$ and $-\sqrt2$ are zeroes, $(x-\sqrt2)(x+\sqrt2)=x^2-2$ is a factor.

Dividing $2x^4-3x^3-3x^2+6x-2$ by $x^2-2$:
$2x^4-3x^3-3x^2+6x-2=(x^2-2)(2x^2-3x+1).$
Now $2x^2-3x+1=(2x-1)(x-1)$, giving zeroes $x=\dfrac{1}{2}$ and $x=1$.

All zeroes: $\sqrt2,\ -\sqrt2,\ 1,\ \dfrac{1}{2}$.

Sum of zeroes: $(a-b)+a+(a+b)=3a=-\dfrac{(-3)}{1}=3\Rightarrow a=1$.

Product of zeroes: $(a-b)(a)(a+b)=a(a^2-b^2)=-\dfrac{1}{1}=-1$.

Substituting $a=1$: $\ 1(1-b^2)=-1\Rightarrow 1-b^2=-1\Rightarrow b^2=2\Rightarrow b=\pm\sqrt2$.

Hence $a=1$ and $b=\pm\sqrt2$.

(i) $y=-x^2+8x-12=-(x-2)(x-6)$, so the zeroes are $x=2$ and $x=6$ (where the ball is on the ground).

(ii) Sum of zeroes $=2+6=8=-\dfrac{b}{a}=-\dfrac{8}{-1}$ ✓; product $=2\times6=12=\dfrac{c}{a}=\dfrac{-12}{-1}$ ✓.

(iii) The ball returns to the ground at $x=6$ m. The maximum height occurs midway at $x=4$, giving $y=-(4)^2+8(4)-12=4$ m.