Question

if x-3 and x-1/3 are both factors of ax 2+5x+b show that a=b

Answer 1

Let take x = 3 ( as x-3 is a factor )
Then the eqn.,
=> 9a + 15 + b = 0 __(1)
& x = 1/3 ( as x-1/3 is also a factor )
Then the eqn.,
=> a/9 + 5/3 + b = 0
=> a + 15 + 9b = 0 __(2)
Solving 1 & 2,
=> a = -3/2, & b = - 3/2. => a = b.
=) Hence proved.

Answer 2

It has been shown that a=b, if x-3 and x-1/3 are both factors of ax²+5x+b.

Given:

• A polynomial $a {x}^{2} + 5x + b$
• Two factors $(x - 3)$ and $x - \frac{1}{3}$

To find:

• Show that a=b.

Solution:

Theorem to be used:

Factors theorem: It states that if a polynomial p(x)(having degree ≥1) have factor (x-a),then p(a)=0.

Step 1:

Put value of x from x-3 in polynomial.

$x - 3 = 0$

Thus,

$x = 3$

Put x=3 to polynomial.

$a( {3)}^{2} + 5(3) + b = 0$

or

$9a + 15 + b = 0$

or

$\bf \pink{9a + b = - 15...eq1}$

Step 2:

Put value of x from x-1/3 in polynomial.

$x - \frac{1}{3} = 0$

or

$x = \frac{1}{3}$

So,

$a {(\frac{1}{3} )}^{2} + 5( \frac{1}{3} ) + b = 0$

or

$\frac{a}{9} + \frac{5}{3} + b = 0$

or

$a + 15 + 9b = 0$

or

$\bf \green{\: a + 9b = - 15...eq2}$

Step 3:

Equate both equations.

As both the equations have RHS equal, so we can equate LHS.

$9a + b = a + 9b$

or

$9a - a = 9b - b$

or

$8a = 8b$

or

$\bf \red{ a = b}$

Thus,

It has been shown that a=b, if x-3 and x-1/3 are both factors of ax²+5x+b.

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