Question

Factorise using factor theorem.
${x}^{4} - 7 {x}^{3} + 9 {x}^{2} + {7x} - 10$
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Answer 1

Question: Factorize $x^4-7x^3+9x^2+7x-10$.

The answer to your question is $(x+1)(x-1)(x-2)(x-5)$.

Solution

By rational root theorem, we know that possible rational zeros are between $\pm1, \pm2,\pm5,\pm10$.

Let's put $x=1$ into our equation.

$\implies (1)^4-7(1)^3+9(1)+7(1)-10=0$

By factor theorem, $x-1$ is a factor.

Let's put $x=-1$ into our equation.

$\implies (-1)^4-7(-1)^3+9(-1)+7(-1)-10=0$

By factor theorem, $x+1$ is a factor.

From above, we know that a factor of the given polynomial is $x^2-1$. Now trying polynomial division to find other factors,

→ Quotient $x^2-7x+10$, remainder 0.

Hence, the polynomial is divided by $x^2-1$ and the other factor is $x^2-7x+10$.

$\implies x^4-7x^3+9x^2+7x-10=(x^2-1)(x^2-7x+10)$

Factorizing completely, we have,

$x^4-7x^3+9x^2+7x-10=(x+1)(x-1)(x-2)(x-5)$

Answer 2

Required Answer :-

$\sf x^4-7x^3+9x^2+7x-10$

$\sf x^4-7x^3+(10x^2-9x^2)+7x-10$

$\sf x^4 - 7x^3+10x^2-9x^2+7x-10$

$\sf x^2\bigg(x^2-7x+10\bigg) - 1\bigg(x^2-7x + 10\bigg)$

$\sf \bigg(x^2-1\bigg)\bigg(x^2-7x+10\bigg)$

Solving them seperately

$\sf x^2-1=0$

$\sf x^2=0+1$

$\sf x^2=1$

$\sf x=\sqrt{1}$

$\sf x =\pm 1$

For second

$\sf x^2-7x+10$

$\sf x^2-\bigg(5x + 2x\bigg) +10$

$\sf x^2-5x -2x+10$

$\sf (x - 5)(x-2)$

So,

Value of x = ±1,5,2