Math, asked by charvi7007, 14 days ago

Without actual division, show that (2x^2 − 7x−15) is a factor of (2x^3– 9x^2 –8x + 15)

Answers

Answered by mathdude500
2

\large\underline{\sf{Solution-}}

Let assume that

\rm \: f(x) =  {2x}^{3} -  {9x}^{2} - 8x + 15 \\

and

\rm \: g(x) =  {2x}^{2} -  7x  -  15 \\

So,

\rm \: g(x) =  {2x}^{2} -  7x  -  15 \\

can be further factorized using splitting of middle terms.

\rm \: g(x) =  {2x}^{2} -  10x + 3x  -  15 \\

\rm \: g(x) =  2x(x - 5)+ 3(x  -  5) \\

\rm \: g(x) =  (x - 5) \: (2x  + 3) \\

Now, in order to show that g(x) is a factor of f(x), we have to show that x - 5 and 2x + 3 is a factor of f(x), using factor theorem.

Factor theorem states that if a polynomial f(x) of degree greater than or equals to 1, then x - a is a factor of f(x) if f(a) = 0.

So, Consider,

\rm \: f(5) \\

\rm \: =  {2(5)}^{3} -  {9(5)}^{2} - 8(5) + 15 \\

\rm \: =  250 -  225 - 40 + 15 \\

\rm \: =  265 -  265  \\

\rm \:  =  \: 0 \\

\rm\implies \:x - 5 \: is \: a \: factor \: of \: f(x) -  -  - (1) \\

Now, Consider

\rm \: f\bigg(-\dfrac{3}{2}\bigg) \\

\rm \:  =  {2\bigg(-\dfrac{3}{2}\bigg)}^{3} -  {9\bigg(-\dfrac{3}{2}\bigg)}^{2} - 8\bigg(-\dfrac{3}{2}\bigg) + 15 \\

\rm \:  =  \:  - \dfrac{27}{4} - \dfrac{81}{4} + 12 + 15 \\

\rm \:  =  \:  \dfrac{ - 27 - 81}{4}  + 27 \\

\rm \:  =  \:   \dfrac{-108}{4}  + 27 \\

\rm \:  =  \:  - 27 + 27 \\

\rm \:  =  \: 0 \\

\rm\implies \:2x + 3 \: is \: a \: factor \: of \: f(x) -  -  - (2) \\

From equation (1) and (2), we concluded that

\rm \: x - 5 \: and \:2x + 3 \: is \: a \: factor \: of \: f(x) \\

\rm\implies \:(x - 5)(2x + 3)\: is \: a \: factor \: of \: f(x) \\

\rm\implies \: {2x}^{2} - 7x - 15 \: is \: a \: factor \: of \: f(x) \\

\rm\implies \: g(x) \: is \: a \: factor \: of \: f(x) \\

\rule{190pt}{2pt}

Basic Concept Used :-

Splitting of middle terms :-

In order to factorize  ax² + bx + c we have to find numbers m and n such that m + n = b and mn = ac.

After finding m and n, we split the middle term i.e bx in the quadratic equation as mx + nx and get the required factors by grouping the terms.

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