A linear polynomial of the form P(x) = ax + b. If k is the zero of P(x), then,

P(k) = ak + b = 0

Zero of the polynomial, k = – b/a = – constant term/coefficient ofx

Now, consider the quadratic polynomial, P(x) = \(4x^2-9x+2\)

Factorisation of P(x) can be done by splitting the middle term into two terms such that their product is a multiple of the first term. ie. multiple of \(4x^2\)

\(-9x\)

\( 4x^2-9x+2\)

= \(4x (x-2)-(x-2)\)

= \((4x-1)(x-2)\)

Zeros of the polynomial \(4x^2-9x+2\)

Zeros are found by equating the polynomial to zero.

i.e.

\( (4x-1)(x-2)\)

Therefore, either \((4x-1)\)

4x – 1 = 0 gives x = \( \frac 14 \)

Zeros of P(x) are \( \frac 14 \)

It is observed that, sum of zeros, \( \frac 14 +2\)

Product of the roots,1/4×2=1/2 = constant term/coefficient of x^2

Let’s take one more example to verify above concept, let P(x) = \(x^2-3x-10\)

To factorise the above polynomial, we have to split the middle term -3x into two terms such that the product of them is a multiple of \(-10x^2\)

Therefore,

-3x can be written as -3x =-5x + 2x, [since -5x × 2x = \(-10x^2]\)

\(x^2-3x-10\)

= x (x – 5) + 2(x – 5)

= (x – 5)(x + 2)

Zeros of P(x) are,

x – 5 = 0, x = 5

x + 2 = 0, x=-2

Sum of zeros,

5 – 2 = \( \frac 31 \)

Product of zeros,

5 × -2 = – \( \frac {10}{1} \)

In general, if α and β are the zeros of the polynomial P(x)= \( ax^2+bx+c,a≠0\)

P(x) can be written as,

\( ax^2+bx+c=k(x-α)(x-β)\)

= \( k[x^2-(α+β)x+αβ]\)

=\(kx^2-(α+β)kx+kαβ\)

Comparing the coefficients of terms gives,

a = k, b = -k(α + β), c = kαβ

It gives,

α + β= – \( \frac bk\)

αβ = \( \frac ck\)

Therefore,

Sum of zeros,

α + β = – b/a = \(– \frac{coefficient\; of\; x}{coefficient \;of\; x^2}\)

Product of zeros,

αβ = c/a =\( \frac {constant\; term}{coefficient\; of\; x^2 }\)

Example: Find a quadratic polynomial whose sum and product of zeros are 7 and 12.

Let α and β be zeros of polynomial of form \(ax^2+bx+c\)

α + β = 7 = \(– \frac ba \)

α + β = 12 = \( \frac ca \)

If a = 1, then b = -7 and c = 12

Therefore, one quadratic polynomial satisfying the above condition is \( x^2-7x+12\)

Now, consider the cubic polynomial P(x) = \(ax^3+bx^2+cx+d\)

α + β + γ = \( – \frac ba \)

αβ + βγ + αγ = \( \frac ca \)

αβγ = \( – \frac da \)

Example: Two zeros of the polynomial P(x) = \(x^3-4x^2+x+6\)

Let the third root be γ,

Comparing polynomial with \(ax^3+bx^2+cx+d\)

a = 1, b = -4, c = 1, d = 6

Sum of zeros of the cubic polynomial

2 – 1 + γ = \(– \frac ba \)

γ = 4 – 1 = 3

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