Question

50 Points ✌



If ${x}^2 - {bx} + {c}$ = $({x} + {p}) ({x} - {q})$ then factorise ${x}^2-{bxy} + {cy}^2$

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Answer 1

We're given that  $x^2-bx+c=(x+p)(x-q).$

We can expand the RHS.

$(x+p)(x-q)=x^2+px-qx-pq=x^2+(p-q)x-pq$

Now,

$x^2-bx+c=x^2+(p-q)x-pq$

From both sides we get,

$-b=p-q\ \ \ \ \ ;\ \ \ \ \ c=-pq$

So,

$x^2-bxy+cy^2\ \Longrightarrow\ x^2+(p-q)xy-(pq)y^2\\ \\ \\ \Longrightarrow\ x^2+pxy-qxy-pqy^2\ \Longrightarrow\ x(x+py)-qy(x+py)\\ \\ \\ \Longrightarrow\ \mathbf{(x+py)(x-qy)}$

Hence factorized!

Or let me show you another method.

Consider  $x^2-bxy+cy^2.$  The value won't change on multiplying this by  $\dfrac{y^2}{y^2}.$

$(x^2-bxy+cy^2 )\ \dfrac{y^2}{y^2}\ \ \Longrightarrow\ \ \dfrac{x^2-bxy+cy^2}{y^2}\times y^2\\ \\ \\ \Longrightarrow\ \left(\dfrac{x^2}{y^2}-\dfrac{bxy}{y^2}+\dfrac{cy^2}{y^2}\right)y^2\ \Longrightarrow\ \left(\left(\dfrac{x}{y}\right)^2-b\left(\dfrac{x}{y}\right)+c\right)y^2$

Now this seems like  $x^2-bx+c.$  except  $y^2.$

If  $x^2-bx+c=(x+p)(x-q),$  then  $\left(\dfrac{x}{y}\right)^2-b\left(\dfrac{x}{y}\right)+c\ =\ \left(\dfrac{x}{y}+p\right)\left(\dfrac{x}{y}-q\right).$

So,

$\left(\left(\dfrac{x}{y}\right)^2-b\left(\dfrac{x}{y}\right)+c\right)y^2\ =\ \left(\dfrac{x}{y}+p\right)\left(\dfrac{x}{y}-q\right)y^2$

Here I split  $y^2$  as  $y\times y,$  so,

$y\left(\dfrac{x}{y}+p\right)\left(\dfrac{x}{y}-q\right)y\ \Longrightarrow\ \mathbf{(x+py)(x-qy)}$

Hence Factorized!

Answer 2

Hey there

refer to attachment