Question

Factorise 10y²+21y-10​

Answer 1

(2y+5)x(5y-2)

Explanation:

Answer 2

For a polynomial of the form ax² - bx + c rewrite the middle term as a sum of two terms whose product is a·c=10·-10 = -100 and whose sum is b= 21.

We factor 21 out of 21y.

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{10y^(2)+21(y)-10 } \end{gathered}$}}`

Rewrite 21 as −4 plus 25.

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{10y^(2)+(-4+25)y-10 } \end{gathered}$}}`

Apply the distributive property.

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{10y^(2)-4y+25y-10 } \end{gathered}$}}`

Factor the highest common denominator of each group. Group the first two terms and the last two.

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{(10y^(2)-4y)+25y-10 } \end{gathered}$}}`

Factor the highest common denominator (GCF) of each group.

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{2y(5y-2)+5(5y-2) } \end{gathered}$}}`

Factor the polynomial by factoring the greatest common denominator, 5y-2.

`\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{(5y-2)(2y+5) } \end{gathered}$}}}`

The answer is: (5y - 2)(2y + 5)