Polynomials — Exercise 2.2

Write each polynomial as az²+bz+c. For zeros α and β, verify α+β=−b/a and αβ=c/a. A repeated zero is used twice in these two relationships.

(i)

\[x^2-2x-8=(x-4)(x+2)\]

The zeros are 4 and −2.

\[4+(-2)=2=\frac{-(-2)}1\]

\[4(-2)=-8=\frac{-8}1\]

(ii)

\[4s^2-4s+1=(2s-1)^2\]

The zero 1/2 is repeated.

\[\frac12+\frac12=1=\frac{-(-4)}4\]

\[\frac12\cdot\frac12=\frac14=\frac14\]

(iii)

\[6x^2-7x-3=6x^2-9x+2x-3\]

\[=3x(2x-3)+(2x-3)=(3x+1)(2x-3)\]

The zeros are −1/3 and 3/2.

\[-\frac13+\frac32=\frac76=\frac{-(-7)}6\]

\[-\frac13\cdot\frac32=-\frac12=\frac{-3}6\]

(iv)

\[4u^2+8u=4u(u+2)\]

The zeros are 0 and −2.

\[0+(-2)=-2=\frac{-8}4\]

\[0(-2)=0=\frac04\]

(v)

\[t^2-15=(t-\sqrt{15})(t+\sqrt{15})\]

The zeros are √15 and −√15.

\[\sqrt{15}+(-\sqrt{15})=0=\frac{-0}1\]

\[\sqrt{15}(-\sqrt{15})=-15=\frac{-15}1\]

(vi)

\[3x^2-x-4=3x^2-4x+3x-4\]

\[=x(3x-4)+(3x-4)=(x+1)(3x-4)\]

The zeros are −1 and 4/3.

\[-1+\frac43=\frac13=\frac{-(-1)}3\]

\[(-1)\frac43=-\frac43=\frac{-4}3\]

If the sum of the zeros is S and their product is P, expand (x−α)(x−β) to obtain x²−Sx+P. Any nonzero constant multiple is equally valid.

(i) S=1/4, P=−1

\[x^2-\frac14x-1\]

Multiplying by 4 gives:

(ii) S=√2, P=1/3

\[x^2-\sqrt2x+\frac13\]

(iii) S=0, P=√5

\[x^2-0x+\sqrt5=x^2+\sqrt5\]

(iv) S=1, P=1

(v) S=−1/4, P=1/4

\[x^2-\left(-\frac14\right)x+\frac14\]

\[=x^2+\frac14x+\frac14\]

(vi) S=4, P=1

Important distinction: Constructing coefficients does not guarantee real zeros. For example, x²−x+1 has discriminant −3, so it has no real zeros. This exercise asks for the quadratic expressions, not for a real-root solution.