Question

Determine the numbers a,b and c so that the polynomials `x+5` and `a(x-2)(x-3)+b(x-1)(x-3)+c(x-1)(x-2)` are identical?

Answer 1

`a=3,b=-7" and "c=4`

Explanation:

`"we require to "color(blue)"compare coefficients of like terms"`

`a(x-2)(x-3)=ax^2-5ax+6a`

`b(x-1)(x-3)=bx^2-4bx+3b`

`c(x-1)(x-2)=cx^2-3cx+2c`

`color(red)"compare with the coefficients of "x+5`

`• " coefficient of "x^2" term is zero"`

`rArra+b+c=0to(1)`

`• " coefficient of x term is 1"`

`rArr-5a-4b-3c=1to(2)`

`• " constant term is 5"`

`rArr6a+3b+2c=5to(3)`

`"now solving the equations using elimination/matrices"`

`"gives "a=3,b=-7" and "c=4`

Answer 2

Answer is `a=3`, `b=-7` and `c=4`

Explanation:

Observe that if we put either `x=1`, `x=2` or `x=3`, two terms in the given expression `a(x-2)(x-3)+b(x-1)(x-3)+c(x-1)(x-2)` become `0`.

Hence to find values of `a,b` and `c`, let us first assume that

`a(x-2)(x-3)+b(x-1)(x-3)+c(x-1)(x-2)=x+5`

then if `x=1`, this becomes `a(1-2)(1-3)=1+5`

i.e. `2a=6` i.e. `a=3`

if `x=2`, this becomes `b(2-1)(2-3)=2+5`

or `-b=7` i.e. `b=-7`

and if `x=3`, then this becomes `c(3-1)(3-2)=3+5`

or `2c=8` i.e. `c=4`

Hence answer is `a=3`, `b=-7` and `c=4`

Putting these values expression becomes

`3(x-2)(x-3)-7(x-1)(x-3)+4(x-1)(x-2)`

= `3(x^2-5x+6)-7(x^2-4x+3)+4(x^2-3x+2)`

= `x+5`