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How do you factor and solve $ - {x^2} - 2x + 3?$

seo-qna
Last updated date: 17th Jun 2024
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Answer
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Hint: As we know that factorising is the reverse of expanding brackets, it is an important way of solving equations. The first step of factoring an expression is to take out any common factors which the terms have. So if we were asked to factor the expression ${x^2} + x$ , since $x$ goes into both terms, we would write $x(x + 1)$. We know that for the factorisation of quadratic polynomials of the form ${x^2} + bx + c$we have to find numbers $p$and $q$such that $p + q = b$ and $pq = c$. This is called factorisation by splitting the middle term.

Complete step-by-step solution:
Here we have an equation: $ - {x^2} - 2x + 3$ .
We will change the sign of the equation i.e. by multiplying the whole equation is negative sign: $ - ( - {x^2} - 2x + 3) = 0$, we get new equation which is ${x^2} + 2x - 3 = 0$.So we have to split the middle term i.e. $(2x)$in such numbers that that the product of the numbers will be equal to $( - 3{x^2})$. We can write $2x = 3x - x$ and $(3x)( - x) = - 3{x^2}$.

Therefore ${x^2} + 2x - 3 = {x^2} + 3x - x - 3$, Now take out the common factor and simplify it;
$x(x + 3) - 1(x + 3) = 0 \Rightarrow (x + 3)(x - 1)$.This is the simplest factor of the given quadratic equation. Now we have $(x + 3) = 0$or $(x - 1) = 0$. So we get two values of $x$ i.e. $x = - 3$ and $x = 1$.

Hence the factors of $ - {x^2} - 2x + 3 = 0$ is and the values of $x$are $ - 3$or $1$.

Note: We should keep in mind while solving this kind of middle term factorisation that we use correct identities to factorise the given algebraic expressions and keep checking the negative and positive sign otherwise it will give the wrong answer. Also we should always check for the sum and product and also verify the factors by multiplying that as it will provide the same above quadratic equation or not. These are some of the standard algebraic identities. This is as far we can go with real coefficients as the remaining quadratic factors all have complex zeros.