Question

factorise x²-4x-21 , x²-9x+18 , x²-19x+78

Answer 1

Concept Introduction:-

It may be in the form of a word, a symbol, or a figure that reflects the arithmetic value of a quantity.

Given Information:-

We have been given that $x^{2}-4x-21 , x^{2}-9x+18 , x^{2}-19x+78$

To Find:-

We have to find that the factorise the value

Solution:-

According to the problem

$x^{2}-4x-21=0\\x^{2}-7x+3x-21=0\\x(x-7)-3(x-7)=0\\(x-7)(x-3)=0\\x=3,7$

$x^2 +9x+18=0\\x^2 +6x+3x+18=0\\x(x+6)+3(x+6)=0\\(x+3)(x+6)=0\\x=-3.-6$

$x^{2}-19x+78=0\\x^{2}-13x-6x+78=0\\x(x-13)-6(x-13)=0\\(x-13)(x-6)=0\\x=13,6$

Final Answer:-

The factors are $(x-7)(x-3),(x+3)(x+6)$ and $(x-13)(x-6)$ respectively.

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